Redução ao 1.º Quadrante | Matemática A

📌 Resumo teórico

Periodicidade

\[ \sin(\alpha+2k\pi)=\sin\alpha\qquad \cos(\alpha+2k\pi)=\cos\alpha\qquad \tan(\alpha+k\pi)=\tan\alpha\qquad (k\in\mathbb{Z}) \]

Fórmulas de redução

	\(-\alpha\)	\(\pi-\alpha\)	\(\pi+\alpha\)	\(2\pi-\alpha\)	\(\frac{\pi}{2}-\alpha\)	\(\frac{\pi}{2}+\alpha\)	\(\frac{3\pi}{2}-\alpha\)	\(\frac{3\pi}{2}+\alpha\)
\(\sin\)	\(-\sin\alpha\)	\(\sin\alpha\)	\(-\sin\alpha\)	\(-\sin\alpha\)	\(\cos\alpha\)	\(\cos\alpha\)	\(-\cos\alpha\)	\(-\cos\alpha\)
\(\cos\)	\(\cos\alpha\)	\(-\cos\alpha\)	\(-\cos\alpha\)	\(\cos\alpha\)	\(\sin\alpha\)	\(-\sin\alpha\)	\(-\sin\alpha\)	\(\sin\alpha\)
\(\tan\)	\(-\tan\alpha\)	\(-\tan\alpha\)	\(\tan\alpha\)	\(-\tan\alpha\)	\(\frac{1}{\tan\alpha}\)	\(-\frac{1}{\tan\alpha}\)	\(\frac{1}{\tan\alpha}\)	\(-\frac{1}{\tan\alpha}\)

Regra prática: com \(\pi\) ou \(2\pi\) a razão mantém-se; com \(\frac{\pi}{2}\) ou \(\frac{3\pi}{2}\) o seno passa a cosseno e vice-versa (e a tangente passa ao inverso). O sinal é o da razão original no quadrante onde fica o ângulo, supondo \(\alpha\) do 1.º quadrante.

As fórmulas com tangentes ou frações aplicam-se apenas onde todas as expressões estão definidas. Nas simplificações, conserva-se o domínio da expressão original.

Valores exatos

\(\alpha\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{3}\)
\(\sin\alpha\)	\(\frac{1}{2}\)	\(\frac{\sqrt2}{2}\)	\(\frac{\sqrt3}{2}\)
\(\cos\alpha\)	\(\frac{\sqrt3}{2}\)	\(\frac{\sqrt2}{2}\)	\(\frac{1}{2}\)
\(\tan\alpha\)	\(\frac{\sqrt3}{3}\)	\(1\)	\(\sqrt3\)

Exemplo: \(\sin\dfrac{5\pi}{6}=\sin\left(\pi-\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{6}=\dfrac{1}{2}\).

1. Simplificação de expressões

Simplifica o mais possível as seguintes expressões.

a) \(\sin(\pi-\alpha)-\sin(2\pi-\alpha)\) \({}+\cos(-\alpha)+\cos(\pi-\alpha)\)
b) \(\tan(\pi-\alpha)-\tan(2\pi-\alpha)\) \({}-3\sin(-\alpha)\) \({}-2\sin(2\pi+\alpha)\)
c) \(\sin(-\alpha)\sin(\pi+\alpha)\) \({}-\cos(-\alpha)\cos(\pi-\alpha)\)
d) \(5\sin\left(\dfrac{\pi}{2}-\alpha\right)\) \({}+\cos\left(\dfrac{\pi}{2}+\alpha\right)\) \({}-\sin(\pi+\alpha)\) \({}+3\cos(\pi+\alpha)\)
e) \(\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \({}+\cos(2\pi-\alpha)\) \({}-\tan(-\alpha)\cos(2\pi-\alpha)\) \({}+\cos\left(\dfrac{3\pi}{2}+\alpha\right)\)
f) \(\sin(\alpha-\pi)-\tan(\alpha+\pi)\) \({}-\tan(2\pi-\alpha)\) \({}-5\cos\left(\alpha-\dfrac{3\pi}{2}\right)\)
g) \(\sin(\alpha-5\pi)\tan(-\alpha-4\pi)\) \({}-3\sin\left(\alpha-\dfrac{\pi}{2}\right)\) \({}+2\cos(7\pi-\alpha)\)
h) \(\cos\left(\dfrac{5\pi}{2}+\alpha\right)\) \({}-\sin\left(\alpha-\dfrac{7\pi}{2}\right)\) \({}-\cos(7\pi-\alpha)\) \({}+2\sin(\alpha+4\pi)\)

\sin(\pi-\alpha)-\sin(2\pi-\alpha)\) \({}+\cos(-\alpha)\) \({}+\cos(\pi-\alpha)=

=\sin\alpha+\sin\alpha\) \({}+\cos\alpha-\cos\alpha\) \(=2\sin\alpha

\tan(\pi-\alpha)-\tan(2\pi-\alpha)\) \({}-3\sin(-\alpha)\) \({}-2\sin(2\pi+\alpha)=

=-\tan\alpha+\tan\alpha\) \({}+3\sin\alpha-2\sin\alpha\) \(=\sin\alpha

\sin(-\alpha)\sin(\pi+\alpha)\) \({}-\cos(-\alpha)\cos(\pi-\alpha)=

=(-\sin\alpha)\times(-\sin\alpha)\) \({}-\cos\alpha\times(-\cos\alpha)\) \(=\sin^2\alpha+\cos^2\alpha=1

5\sin\left(\dfrac{\pi}{2}-\alpha\right)\) \({}+\cos\left(\dfrac{\pi}{2}+\alpha\right)\) \({}-\sin(\pi+\alpha)\) \({}+3\cos(\pi+\alpha)=

=5\cos\alpha-\sin\alpha\) \({}+\sin\alpha-3\cos\alpha\) \(=2\cos\alpha

\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \({}+\cos(2\pi-\alpha)\) \({}-\tan(-\alpha)\cos(2\pi-\alpha)\) \({}+\cos\left(\dfrac{3\pi}{2}+\alpha\right)=

=-\cos\alpha+\cos\alpha\) \({}+\tan\alpha\times\cos\alpha+\sin\alpha\) \(=\dfrac{\sin\alpha}{\cos\alpha}\times\cos\alpha+\sin\alpha\) \(=2\sin\alpha

\sin(\alpha-\pi)-\tan(\alpha+\pi)\) \({}-\tan(2\pi-\alpha)\) \({}-5\cos\left(\alpha-\dfrac{3\pi}{2}\right)=

=-\sin\alpha-\tan\alpha\) \({}+\tan\alpha+5\sin\alpha\) \(=4\sin\alpha

\sin(\alpha-5\pi)\) \({}\times\tan(-\alpha-4\pi)\) \({}-3\sin\left(\alpha-\dfrac{\pi}{2}\right)\) \({}+2\cos(7\pi-\alpha)=

=-\sin\alpha\times(-\tan\alpha)\) \({}+3\cos\alpha-2\cos\alpha\) \(=\sin\alpha\times\dfrac{\sin\alpha}{\cos\alpha}\) \({}+\cos\alpha=

=\dfrac{\sin^2\alpha}{\cos\alpha}+\cos\alpha\) \(=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha}\) \(=\dfrac{1}{\cos\alpha}

\cos\left(\dfrac{5\pi}{2}+\alpha\right)\) \({}-\sin\left(\alpha-\dfrac{7\pi}{2}\right)\) \({}-\cos(7\pi-\alpha)\) \({}+2\sin(\alpha+4\pi)=

=-\sin\alpha-\cos\alpha\) \({}+\cos\alpha+2\sin\alpha\) \(=\sin\alpha

a) \(2\sin\alpha\); b) \(\sin\alpha\); c) \(1\); d) \(2\cos\alpha\);
e) \(2\sin\alpha\); f) \(4\sin\alpha\); g) \(\dfrac{1}{\cos\alpha}\); h) \(\sin\alpha\).

2. Valor lógico de uma proposição

Qual é o valor lógico da proposição:

Se \(\sin(\pi-x)>0\) e \(\tan\left(-\dfrac{\pi}{2}-x\right)<0\), então \(\cos(-\pi-x)<0\).

Estudo das condições

\sin(\pi-x)>0\Leftrightarrow\sin x>0

\tan\left(-\dfrac{\pi}{2}-x\right)<0\Leftrightarrow\tan x<0

Sabemos que a tangente é negativa nos 2.º e 4.º quadrantes. Como \(\sin x>0\), o lado extremidade de \(x\) está no 2.º quadrante.

Estudo da conclusão

\cos(-\pi-x)\) \(=\cos\big(-(\pi+x)\big)\) \(=\cos(\pi+x)\) \(=-\cos x

No 2.º quadrante, o cosseno é negativo, então:

\cos x<0\Rightarrow-\cos x>0\Rightarrow\cos(-\pi-x)>0

Portanto, a proposição é falsa: as hipóteses são satisfeitas, mas a conclusão tem o sinal contrário. Por exemplo, para \(x=\frac{3\pi}{4}\), as duas hipóteses são verdadeiras e \(\cos(-\pi-x)=\frac{\sqrt2}{2}>0\).

3. Demonstração de identidades

Prova que, para qualquer \(x\) para o qual as expressões têm significado, se tem:

a) \(1-\dfrac{\cos(2\pi+x)}{\tan(2\pi+x)\sin x+\cos(2\pi-x)}\) \(=\sin^2 x\)
b) \(2\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}+x\right)\) \({}-\cos(2\pi)\) \(=\left[1-\sqrt{2}\,\sin(2\pi+x)\right]\) \(\left[1-\sqrt{2}\,\sin(\pi+x)\right]\)

1-\dfrac{\cos(2\pi+x)}{\tan(2\pi+x)\sin x+\cos(2\pi-x)}\) \(=1-\dfrac{\cos x}{\tan x\times\sin x+\cos x}=

=1-\dfrac{\cos x}{\dfrac{\sin x}{\cos x}\times\sin x+\cos x}\) \(=1-\dfrac{\cos x}{\dfrac{\sin^2 x+\cos^2 x}{\cos x}}\) \(=1-\dfrac{\cos x}{\dfrac{1}{\cos x}}\) \(=1-\cos^2 x\) \(=\sin^2 x

2\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}+x\right)\) \({}-\cos(2\pi)\) \(=2\cos x\cos x\) \({}-\cos(2\pi)\) \(=2\cos^2 x-1=

=2(1-\sin^2 x)-1\) \(=2-2\sin^2 x-1\) \(=1-2\sin^2 x\) \(=1-\left(\sqrt{2}\,\sin x\right)^2=

=\left(1-\sqrt{2}\,\sin x\right)\left(1+\sqrt{2}\,\sin x\right)\) \(=\left(1-\sqrt{2}\,\sin(2\pi+x)\right)\) \(\left(1-\sqrt{2}\,\sin(\pi+x)\right)

Em ambos os casos as identidades ficam demonstradas. \(\blacksquare\)

4. Valor de expressões

Determina o valor das seguintes expressões.

a) \(\sin\left(-\dfrac{\pi}{6}\right)\) \({}+\sqrt{6}\,\sin\left(-\dfrac{\pi}{4}\right)\sin\left(-\dfrac{\pi}{3}\right)\)
b) \(\sin\left(-\dfrac{3\pi}{4}\right)-\cos\left(-\dfrac{\pi}{4}\right)\) \({}+\sqrt{5-\sqrt{3}\,\tan\left(-\dfrac{\pi}{3}\right)}\)
c) \(\dfrac{\sin\dfrac{5\pi}{6}+\cos\left(-\dfrac{3\pi}{4}\right)}{2+\cos(-\pi)+\cos\left(-\dfrac{2\pi}{3}\right)-\sin\left(-\dfrac{3\pi}{4}\right)}\)
d) \(\dfrac{\sin\left(-\dfrac{17\pi}{3}\right)}{\cos\dfrac{17\pi}{4}}+\sqrt{6}\,\sin\dfrac{37\pi}{6}\)

\sin\left(-\dfrac{\pi}{6}\right)\) \({}+\sqrt{6}\,\sin\left(-\dfrac{\pi}{4}\right)\sin\left(-\dfrac{\pi}{3}\right)\) \(=-\sin\dfrac{\pi}{6}\) \({}+\left(-\sqrt{6}\,\sin\dfrac{\pi}{4}\right)\left(-\sin\dfrac{\pi}{3}\right)=

=-\sin\dfrac{\pi}{6}\) \({}+\sqrt{6}\,\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{3}\) \(=-\dfrac{1}{2}+\sqrt{6}\times\dfrac{\sqrt{2}}{2}\times\dfrac{\sqrt{3}}{2}\) \(=-\dfrac{1}{2}+\dfrac{6}{4}\) \(=-\dfrac{1}{2}+\dfrac{3}{2}=1

\sin\left(-\dfrac{3\pi}{4}\right)\) \({}-\cos\left(-\dfrac{\pi}{4}\right)\) \({}+\sqrt{5-\sqrt{3}\,\tan\left(-\dfrac{\pi}{3}\right)}\) \(=-\sin\dfrac{\pi}{4}-\cos\dfrac{\pi}{4}\) \({}+\sqrt{5+\sqrt{3}\,\tan\dfrac{\pi}{3}}=

=-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\) \({}+\sqrt{5+\sqrt{3}\times\sqrt{3}}\) \(=-\sqrt{2}+\sqrt{8}\) \(=-\sqrt{2}+2\sqrt{2}=\sqrt{2}

\dfrac{\sin\dfrac{5\pi}{6}+\cos\left(-\dfrac{3\pi}{4}\right)}{2+\cos(-\pi)+\cos\left(-\dfrac{2\pi}{3}\right)-\sin\left(-\dfrac{3\pi}{4}\right)}\) \(=\dfrac{\sin\dfrac{\pi}{6}-\cos\dfrac{\pi}{4}}{2-1-\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{4}}\) \(=\dfrac{\dfrac{1}{2}-\dfrac{\sqrt{2}}{2}}{1-\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}}=

=\dfrac{\dfrac{1-\sqrt{2}}{2}}{\dfrac{1+\sqrt{2}}{2}}\) \(=\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\) \(=\dfrac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\) \(=\dfrac{1-2\sqrt{2}+2}{1-2}\) \(=\dfrac{3-2\sqrt{2}}{-1}\) \(=2\sqrt{2}-3

\dfrac{\sin\left(-\dfrac{17\pi}{3}\right)}{\cos\dfrac{17\pi}{4}}\) \({}+\sqrt{6}\,\sin\dfrac{37\pi}{6}\) \(=\dfrac{\sin\left(-\dfrac{17\pi}{3}+6\pi\right)}{\cos\left(\dfrac{17\pi}{4}-4\pi\right)}\) \({}+\sqrt{6}\,\sin\left(\dfrac{37\pi}{6}-6\pi\right)=

=\dfrac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{4}}+\sqrt{6}\,\sin\dfrac{\pi}{6}\) \(=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{\sqrt{2}}{2}}+\sqrt{6}\times\dfrac{1}{2}\) \(=\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{6}}{2}\) \(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{2}=\sqrt{6}

a) \(1\); b) \(\sqrt{2}\); c) \(2\sqrt{2}-3\); d) \(\sqrt{6}\).

5. Determinação de x e cálculo

Seja \(x\in\left[\dfrac{3\pi}{2},\,2\pi\right]\), tal que \(\cos\left(\dfrac{\pi}{2}-x\right)\tan(\pi+x)\) \({}-\sin\left(\dfrac{3\pi}{2}-x\right)=2\).

a) Determina o valor de \(x\).
b) Determina o valor de \(\tan(\pi-x)+2\cos(\pi+x)\) \({}-2\sin(2\pi-x)\).

a) Determinação de x

\cos\left(\dfrac{\pi}{2}-x\right)\tan(\pi+x)\) \({}-\sin\left(\dfrac{3\pi}{2}-x\right)=2\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow

\Leftrightarrow\sin x\times\tan x\) \({}+\cos x=2\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow

\Leftrightarrow\sin x\times\dfrac{\sin x}{\cos x}\) \({}+\cos x=2\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow

\Leftrightarrow\dfrac{\sin^2 x+\cos^2 x}{\cos x}=2\) \(\Leftrightarrow\dfrac{1}{\cos x}=2\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow

\Leftrightarrow 1=2\cos x\) \({}\wedge\ \cos x\neq0\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow

\Leftrightarrow\cos x=\dfrac{1}{2}\) \({}\wedge\ x\in\left[\dfrac{3\pi}{2},2\pi\right]\) \(\Leftrightarrow x=\dfrac{5\pi}{3}

a) \(x=\dfrac{5\pi}{3}\).

b) Cálculo da expressão (com \(x=\dfrac{5\pi}{3}\))

\tan\left(\pi-\dfrac{5\pi}{3}\right)\) \({}+2\cos\left(\pi+\dfrac{5\pi}{3}\right)\) \({}-2\sin\left(2\pi-\dfrac{5\pi}{3}\right)=

=\tan\left(-\dfrac{2\pi}{3}\right)\) \({}+2\cos\dfrac{8\pi}{3}-2\sin\dfrac{\pi}{3}\) \(=\tan\dfrac{\pi}{3}+2\cos\dfrac{2\pi}{3}\) \({}-2\sin\dfrac{\pi}{3}=

=\sqrt{3}-2\cos\dfrac{\pi}{3}\) \({}-2\times\dfrac{\sqrt{3}}{2}\) \(=\sqrt{3}-2\) \({}\times\dfrac{1}{2}-\sqrt{3}=-1

b) O valor da expressão é \(-1\).

6. Valores exatos em radianos I

Calcule o valor exato de:

6.1. \(\cos\left(\dfrac{5\pi}{4}\right)+\sin\left(-\dfrac{3\pi}{4}\right)\)

6.2. \(\dfrac{\cos\left(-\frac{7\pi}{6}\right)+\sin\left(\frac{5\pi}{3}\right)}{\tan\left(-\frac{\pi}{3}\right)}\)

6.3. \(\cos\left(\dfrac{\pi}{18}\right)+\cos\left(-\dfrac{17\pi}{18}\right)\) \({}-\cos\left(-\dfrac{5\pi}{6}\right)\)

6.4. \(\dfrac{\tan\left(-\frac{\pi}{4}\right)+\sin^2\left(-\frac{3\pi}{4}\right)}{\cos\left(-\frac{5\pi}{6}\right)}\)

6.1

\cos\left(\pi+\dfrac{\pi}{4}\right)\) \({}-\sin\left(\pi-\dfrac{\pi}{4}\right)\) \(=-\cos\dfrac{\pi}{4}-\sin\dfrac{\pi}{4}\) \(=-\dfrac{\sqrt2}{2}-\dfrac{\sqrt2}{2}\) \(=-\sqrt2

6.2

\cos\left(-\dfrac{7\pi}{6}\right)\) \(=\cos\left(\pi+\dfrac{\pi}{6}\right)\) \(=-\dfrac{\sqrt3}{2}

\sin\dfrac{5\pi}{3}\) \(=\sin\left(2\pi-\dfrac{\pi}{3}\right)\) \(=-\dfrac{\sqrt3}{2}

\tan\left(-\dfrac{\pi}{3}\right)=-\sqrt3

\dfrac{-\frac{\sqrt3}{2}-\frac{\sqrt3}{2}}{-\sqrt3}=\dfrac{-\sqrt3}{-\sqrt3}=1

6.3

\cos\left(-\dfrac{17\pi}{18}\right)\) \(=\cos\left(\pi-\dfrac{\pi}{18}\right)\) \(=-\cos\dfrac{\pi}{18}

\cos\left(-\dfrac{5\pi}{6}\right)\) \(=-\cos\dfrac{\pi}{6}\) \(=-\dfrac{\sqrt3}{2}

\cos\dfrac{\pi}{18}-\cos\dfrac{\pi}{18}+\dfrac{\sqrt3}{2}\) \(=\dfrac{\sqrt3}{2}

6.4

\dfrac{-1+\left(-\frac{\sqrt2}{2}\right)^2}{-\frac{\sqrt3}{2}}\) \(=\dfrac{-\frac12}{-\frac{\sqrt3}{2}}\) \(=\dfrac{1}{\sqrt3}\) \(=\dfrac{\sqrt3}{3}

6.1. \(-\sqrt2\)
6.2. \(1\)
6.3. \(\dfrac{\sqrt3}{2}\)
6.4. \(\dfrac{\sqrt3}{3}\)

7. Valores exatos em radianos II

Determine o valor exato de cada uma das seguintes expressões.

7.1. \(\sin\left(-\dfrac{5\pi}{3}\right)-\cos\dfrac{5\pi}{6}\) \({}+\tan\dfrac{2\pi}{3}-\sin\dfrac{7\pi}{6}\)

7.2. \(\cos\dfrac{11\pi}{6}\times\tan\left(-\dfrac{3\pi}{4}\right)\) \({}-\sin\left(-\dfrac{4\pi}{3}\right)+\cos\dfrac{3\pi}{4}\)

7.3. \(\tan\dfrac{29\pi}{3}+\sin\dfrac{27\pi}{4}\) \({}+\sin\left(-\dfrac{31\pi}{6}\right)+\cos\dfrac{40\pi}{3}\)

7.4. \(\cos\dfrac{150\pi}{4}-\tan\dfrac{123\pi}{3}\) \({}+\cos\dfrac{139\pi}{6}\times\tan\dfrac{100\pi}{3}\)

7.1

\sin\left(-\dfrac{5\pi}{3}+2\pi\right)\) \(=\sin\dfrac{\pi}{3}\) \(=\dfrac{\sqrt3}{2}

\cos\dfrac{5\pi}{6}\) \(=-\dfrac{\sqrt3}{2}

\tan\dfrac{2\pi}{3}\) \(=-\sqrt3

\sin\dfrac{7\pi}{6}\) \(=-\dfrac12

\dfrac{\sqrt3}{2}+\dfrac{\sqrt3}{2}-\sqrt3+\dfrac12=\dfrac12

7.2

\cos\dfrac{11\pi}{6}\) \(=\dfrac{\sqrt3}{2}

\tan\left(-\dfrac{3\pi}{4}\right)\) \(=\tan\dfrac{\pi}{4}=1

\sin\left(-\dfrac{4\pi}{3}\right)\) \(=\sin\dfrac{2\pi}{3}\) \(=\dfrac{\sqrt3}{2}

\cos\dfrac{3\pi}{4}\) \(=-\dfrac{\sqrt2}{2}

\dfrac{\sqrt3}{2}\times1-\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\) \(=-\dfrac{\sqrt2}{2}

7.3 - Retirar voltas (a tangente tem período π)

\dfrac{29\pi}{3}=9\pi+\dfrac{2\pi}{3}

\dfrac{27\pi}{4}=6\pi+\dfrac{3\pi}{4}

-\dfrac{31\pi}{6}\) \(=-6\pi+\dfrac{5\pi}{6}

\dfrac{40\pi}{3}=12\pi+\dfrac{4\pi}{3}

\tan\dfrac{2\pi}{3}+\sin\dfrac{3\pi}{4}\) \({}+\sin\dfrac{5\pi}{6}+\cos\dfrac{4\pi}{3}\) \(=-\sqrt3+\dfrac{\sqrt2}{2}+\dfrac12-\dfrac12\) \(=-\sqrt3+\dfrac{\sqrt2}{2}

7.4

\dfrac{150\pi}{4}=36\pi+\dfrac{3\pi}{2}

\dfrac{123\pi}{3}=41\pi

\dfrac{139\pi}{6}=22\pi+\dfrac{7\pi}{6}

\dfrac{100\pi}{3}=33\pi+\dfrac{\pi}{3}

\cos\dfrac{3\pi}{2}-\tan0\) \({}+\cos\dfrac{7\pi}{6}\times\tan\dfrac{\pi}{3}\) \(=0-0-\dfrac{\sqrt3}{2}\times\sqrt3\) \(=-\dfrac32

7.1. \(\dfrac12\)
7.2. \(-\dfrac{\sqrt2}{2}\)
7.3. \(-\sqrt3+\dfrac{\sqrt2}{2}\)
7.4. \(-\dfrac32\)

8. Sem recorrer à calculadora

Sem recorrer à calculadora, determine o valor das seguintes expressões.

8.1. \(\sin\left(-\dfrac{\pi}{4}\right)+\cos\left(\dfrac{5\pi}{6}\right)\) \({}\times\sin\left(-\dfrac{\pi}{2}\right)\)

8.2. \(\cos(150^\circ)+\sin(300^\circ)-\cos(-120^\circ)\)

8.3. \(\dfrac{\tan\left(-\frac{5\pi}{4}\right)}{\sin\left(\frac{7\pi}{6}\right)}+\cos\left(-\dfrac{8\pi}{3}\right)\) \({}+\sin(-8\pi)\)

8.4. \(\sin\left(-\dfrac{5\pi}{4}\right)+\cos\left(-\dfrac{9\pi}{6}\right)\) \({}-\cos\left(\dfrac{7\pi}{3}\right)\)

8.1\colon\) \(-\dfrac{\sqrt2}{2}+\left(-\dfrac{\sqrt3}{2}\right)\times(-1)\) \(=\dfrac{\sqrt3-\sqrt2}{2}

8.2\colon\) \(-\cos30^\circ-\sin60^\circ\) \({}-\cos120^\circ\) \(=-\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{2}+\dfrac12\) \(=\dfrac12-\sqrt3

8.3\colon\) \(\tan\left(-\dfrac{5\pi}{4}\right)\) \(=-\tan\dfrac{\pi}{4}=-1

\sin\dfrac{7\pi}{6}\) \(=-\dfrac12

\cos\dfrac{8\pi}{3}\) \(=\cos\dfrac{2\pi}{3}\) \(=-\dfrac12

\sin(-8\pi)=0

\dfrac{-1}{-\frac12}-\dfrac12+0\) \(=2-\dfrac12=\dfrac32

8.4\colon\) \(\sin\left(-\dfrac{5\pi}{4}\right)\) \(=-\sin\dfrac{5\pi}{4}\) \(=\dfrac{\sqrt2}{2}

\cos\left(-\dfrac{3\pi}{2}\right)=0

\cos\dfrac{7\pi}{3}\) \(=\cos\dfrac{\pi}{3}=\dfrac12

\dfrac{\sqrt2}{2}+0-\dfrac12=\dfrac{\sqrt2-1}{2}

8.1. \(\dfrac{\sqrt3-\sqrt2}{2}\)
8.2. \(\dfrac12-\sqrt3\)
8.3. \(\dfrac32\)
8.4. \(\dfrac{\sqrt2-1}{2}\)

9. Valores exatos em radianos III

Determine o valor exato das seguintes expressões, considerando as amplitudes em radianos.

9.1. \(\sin\dfrac{19\pi}{6}-\cos\dfrac{11\pi}{3}+\tan\dfrac{21\pi}{4}\)

9.2. \(\tan\dfrac{11\pi}{4}-\sin\dfrac{15\pi}{2}+\cos\dfrac{29\pi}{4}\)

9.1

\dfrac{19\pi}{6}=2\pi+\dfrac{7\pi}{6}

\dfrac{11\pi}{3}=4\pi-\dfrac{\pi}{3}

\dfrac{21\pi}{4}=4\pi+\dfrac{5\pi}{4}

\sin\dfrac{7\pi}{6}-\cos\left(-\dfrac{\pi}{3}\right)\) \({}+\tan\dfrac{5\pi}{4}\) \(=-\dfrac12-\dfrac12+1=0

9.2

\dfrac{11\pi}{4}=2\pi+\dfrac{3\pi}{4}

\dfrac{15\pi}{2}=6\pi+\dfrac{3\pi}{2}

\dfrac{29\pi}{4}=6\pi+\dfrac{5\pi}{4}

\tan\dfrac{3\pi}{4}-\sin\dfrac{3\pi}{2}\) \({}+\cos\dfrac{5\pi}{4}\) \(=-1-(-1)-\dfrac{\sqrt2}{2}\) \(=-\dfrac{\sqrt2}{2}

9.1. \(0\); 9.2. \(-\dfrac{\sqrt2}{2}\)

10. Valores exatos em graus

Determine o valor exato de cada uma das seguintes expressões.

10.1. \(\cos225^\circ+\sin120^\circ+\cos330^\circ-\sin315^\circ\)10.2. \(\sin210^\circ+\cos150^\circ\times\tan135^\circ-\cos300^\circ\)10.3. \(\cos1305^\circ\times\sin1590^\circ\times\tan780^\circ\)10.4. \(\tan1575^\circ+\cos765^\circ\times\sin1395^\circ\)

10.1\colon\) \(-\dfrac{\sqrt2}{2}+\dfrac{\sqrt3}{2}+\dfrac{\sqrt3}{2}\) \({}-\left(-\dfrac{\sqrt2}{2}\right)=\sqrt3

10.2\colon\) \(-\dfrac12+\left(-\dfrac{\sqrt3}{2}\right)\) \({}\times(-1)-\dfrac12\) \(=\dfrac{\sqrt3}{2}-1

10.3\colon\) \(1305^\circ=225^\circ+3\) \({}\times360^\circ;\ 1590^\circ\) \(=150^\circ+4\times360^\circ;\ 780^\circ\) \(=60^\circ+2\times360^\circ

\cos225^\circ\times\sin150^\circ\) \({}\times\tan60^\circ\) \(=-\dfrac{\sqrt2}{2}\times\dfrac12\times\sqrt3\) \(=-\dfrac{\sqrt6}{4}

10.4\colon\) \(\tan135^\circ+\cos45^\circ\) \({}\times\sin315^\circ\) \(=-1+\dfrac{\sqrt2}{2}\times\left(-\dfrac{\sqrt2}{2}\right)\) \(=-\dfrac32

10.1. \(\sqrt3\)
10.2. \(\dfrac{\sqrt3}{2}-1\)
10.3. \(-\dfrac{\sqrt6}{4}\)
10.4. \(-\dfrac32\)

11. Tangente de 17π/6

Qual é o valor exato de \(\tan\left(\dfrac{17\pi}{6}\right)\)?

(A) \(-\dfrac{\sqrt3}{3}\)(B) \(\dfrac{\sqrt3}{3}\)(C) \(-\sqrt3\)(D) \(\sqrt3\)

\tan\dfrac{17\pi}{6}\) \(=\tan\left(2\pi+\pi-\dfrac{\pi}{6}\right)\) \(=-\tan\dfrac{\pi}{6}\) \(=-\dfrac{\sqrt3}{3}

Opção (A).

12. Demonstração de identidades II

Mostre que:

12.1. \(\cos(\pi-\alpha)-2\sin(\alpha-\pi)\) \({}+\cos\left(\dfrac{\pi}{2}-\alpha\right)\) \(=3\sin\alpha-\cos\alpha\)

12.2. \(\tan(3\pi+\alpha)+\dfrac{\cos\left(\frac{\pi}{2}+\alpha\right)}{\sin\left(\frac{\pi}{2}+\alpha\right)}\) \({}-\tan(-\pi-\alpha)\) \(=\tan\alpha\)

12.3. \(\sin\left(\dfrac{\pi}{2}-\alpha\right)+\cos(\alpha-3\pi)\) \({}+2\sin\left(\alpha-\dfrac{\pi}{2}\right)\) \({}+3\sin(\pi+\alpha)\times\tan\left(\dfrac{\pi}{2}-\alpha\right)\) \(=-5\cos\alpha\)

12.1

\(\cos(\pi-\alpha)=-\cos\alpha\), \(\sin(\alpha-\pi)\) \(=-\sin(\pi-\alpha)\) \(=-\sin\alpha\) e \(\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha\):

-\cos\alpha\) \({}-2(-\sin\alpha)+\sin\alpha\) \(=3\sin\alpha-\cos\alpha

12.2

\(\tan(3\pi+\alpha)=\tan\alpha\), \(\cos\left(\frac{\pi}{2}+\alpha\right)=-\sin\alpha\), \(\sin\left(\frac{\pi}{2}+\alpha\right)=\cos\alpha\) e \(\tan(-\pi-\alpha)\) \(=-\tan(\pi+\alpha)\) \(=-\tan\alpha\):

\tan\alpha+\dfrac{-\sin\alpha}{\cos\alpha}\) \({}-(-\tan\alpha)\) \(=\tan\alpha-\tan\alpha+\tan\alpha\) \(=\tan\alpha

12.3

\(\sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha\); \(\cos(\alpha-3\pi)\) \(=\cos(\alpha-\pi)\) \(=-\cos\alpha\); \(\sin\left(\alpha-\frac{\pi}{2}\right)=-\cos\alpha\); \(\sin(\pi+\alpha)=-\sin\alpha\); \(\tan\left(\frac{\pi}{2}-\alpha\right)=\frac{\cos\alpha}{\sin\alpha}\):

\cos\alpha-\cos\alpha-2\cos\alpha\) \({}+3(-\sin\alpha)\times\dfrac{\cos\alpha}{\sin\alpha}\) \(=-2\cos\alpha-3\cos\alpha\) \(=-5\cos\alpha

13. Simplificação de expressões II

Simplifique cada uma das seguintes expressões.

13.1. \(\tan(\pi+\alpha)\times\sin\left(-\dfrac{\pi}{2}+\alpha\right)\) \({}-\cos\left(\dfrac{5\pi}{2}-\alpha\right)\)

13.2. \(\dfrac{\sin(3\pi+\alpha)}{\cos(-8\pi-\alpha)}\) \({}-\tan(\alpha-10\pi)\) \({}+\sin\left(-\dfrac{3\pi}{2}+\alpha\right)\)

13.3. \(\tan(-\alpha)+\sin\left(\dfrac{5\pi}{2}+\alpha\right)\) \({}+\cos(-3\pi-\alpha)\)

13.1

\tan\alpha\times(-\cos\alpha)-\sin\alpha\) \(=-\dfrac{\sin\alpha}{\cos\alpha}\) \({}\times\cos\alpha-\sin\alpha\) \(=-2\sin\alpha

13.2

\dfrac{-\sin\alpha}{\cos\alpha}-\tan\alpha+\cos\alpha\) \(=-2\tan\alpha+\cos\alpha

13.3

-\tan\alpha+\cos\alpha\) \({}+\cos(\pi+\alpha)\) \(=-\tan\alpha\) \({}+\cos\alpha-\cos\alpha\) \(=-\tan\alpha

13.1. \(-2\sin\alpha\)
13.2. \(\cos\alpha-2\tan\alpha\)
13.3. \(-\tan\alpha\)

14. Simplificação de expressões III

No domínio em que têm significado, simplifique cada uma das seguintes expressões.

14.1. \(\cos\left(\dfrac{\pi}{2}-x\right)+\cos\left(\dfrac{\pi}{2}+x\right)\) \({}-\sin(\pi+x)\)

14.2. \(\sin\left(-\dfrac{3\pi}{2}+x\right)\) \({}+\tan(5\pi+x)\) \({}-\sin\left(-\dfrac{\pi}{2}-x\right)\)

14.3. \(\tan\left(x-\dfrac{\pi}{2}\right)+\tan(3\pi-x)\) \({}+\tan\left(\dfrac{7\pi}{2}-x\right)\)

14.4. \(\sin\left(\dfrac{11\pi}{2}-x\right)\) \({}-\cos\left(\dfrac{23\pi}{2}-x\right)\) \({}-\sin(29\pi-x)\)

14.5. \(\tan(x-\pi)\) \({}+\cos\left(-\dfrac{17\pi}{2}+x\right)\) \({}+\tan(x-12\pi)\)

14.6. \(\sin\left(\dfrac{32\pi}{2}+x\right)\) \({}-\sin\left(x-\dfrac{51\pi}{2}\right)\) \({}-\cos(-x-125\pi)\)

14.1\colon\) \(\sin x-\sin x-(-\sin x)\) \(=\sin x

14.2\colon\) \(\sin\left(x+\dfrac{\pi}{2}\right)+\tan x\) \({}+\sin\left(\dfrac{\pi}{2}+x\right)\) \(=\cos x+\tan x+\cos x\) \(=2\cos x+\tan x

14.3\colon\) \(\tan\left(x-\dfrac{\pi}{2}\right)\) \(=-\dfrac{1}{\tan x}

\tan(3\pi-x)\) \(=-\tan x

\tan\left(\dfrac{7\pi}{2}-x\right)\) \(=\tan\left(\dfrac{\pi}{2}-x\right)\) \(=\dfrac{1}{\tan x}

-\dfrac{1}{\tan x}-\tan x+\dfrac{1}{\tan x}\) \(=-\tan x

14.4\colon\) \(\sin\left(\dfrac{3\pi}{2}-x\right)\) \({}-\cos\left(\dfrac{3\pi}{2}-x\right)\) \({}-\sin(\pi-x)\) \(=-\cos x\) \({}+\sin x-\sin x\) \(=-\cos x

14.5\colon\) \(\tan x+\cos\left(x-\dfrac{\pi}{2}\right)+\tan x\) \(=2\tan x+\sin x

14.6\colon\) \(\sin(16\pi+x)\) \({}-\sin\left(x+\dfrac{\pi}{2}\right)\) \({}-\cos(x+\pi)\) \(=\sin x-\cos x+\cos x\) \(=\sin x

14.1. \(\sin x\)
14.2. \(2\cos x+\tan x\)
14.3. \(-\tan x\)
14.4. \(-\cos x\)
14.5. \(2\tan x+\sin x\)
14.6. \(\sin x\)

15. Simplificação de expressões em graus

Simplifique cada uma das seguintes expressões.

15.1. \(\sin(450^\circ+x)-\tan(900^\circ+x)\) \({}+\sin(-90^\circ-x)\)

15.2. \(\cos(x-90^\circ)+\sin(180^\circ+x)\) \({}+\cos(180^\circ-x)\)

15.3. \(\tan(450^\circ-x)-\tan(180^\circ-x)\) \({}-\sin(270^\circ-x)\) \({}-\tan(-270^\circ-x)\)

15.4. \(\tan(-x-90^\circ)\) \({}+\tan(450^\circ+x)\) \({}+\tan(x-180^\circ)\)

15.5. \(-\cos(270^\circ+x)\) \({}+\sin(x-180^\circ)\) \({}-\cos(-x-180^\circ)\)

15.1\colon\) \(\sin(90^\circ+x)-\tan x\) \({}-\sin(90^\circ+x)\) \(=\cos x-\tan x-\cos x\) \(=-\tan x

15.2\colon\) \(\sin x-\sin x-\cos x\) \(=-\cos x

15.3\colon\) \(\tan(90^\circ-x)+\tan x\) \({}+\cos x-\tan(90^\circ-x)\) \(=\tan x+\cos x

15.4\colon\) \(-\tan(x+90^\circ)\) \({}+\tan(90^\circ+x)+\tan x\) \(=\tan x

15.5\colon\) \(-\sin x-\sin x\) \({}-\cos(x+180^\circ)\) \(=-2\sin x+\cos x

15.1. \(-\tan x\)
15.2. \(-\cos x\)
15.3. \(\tan x+\cos x\)
15.4. \(\tan x\)
15.5. \(\cos x-2\sin x\)

16. Exprimir em função das razões de α

Exprima em função das razões trigonométricas de \(\alpha\), em radianos, as seguintes expressões:

16.1. \(\sin(\pi+\alpha)+\cos\left(\dfrac{\pi}{2}-\alpha\right)-\tan(-\alpha)\)16.2. \(\sin(8\pi-\alpha)-\cos(5\pi+\alpha)\)16.3. \(\cos\left(\dfrac{\pi}{2}+\alpha\right)-\sin(-\alpha)-\tan(\pi-\alpha)\)16.4. \(\cos\left(\dfrac{3\pi}{2}-\alpha\right)-\sin(5\pi+\alpha)\)

16.1\colon\) \(-\sin\alpha+\sin\alpha+\tan\alpha\) \(=\tan\alpha

16.2\colon\) \(\sin(-\alpha)-\cos(\pi+\alpha)\) \(=-\sin\alpha+\cos\alpha

16.3\colon\) \(-\sin\alpha+\sin\alpha+\tan\alpha\) \(=\tan\alpha

16.4\colon\) \(-\sin\alpha-\sin(\pi+\alpha)\) \(=-\sin\alpha+\sin\alpha=0

16.1. \(\tan\alpha\)
16.2. \(\cos\alpha-\sin\alpha\)
16.3. \(\tan\alpha\)
16.4. \(0\)

17. A expressão A(x)

Considere a expressão \(A(x)=(\sin x-\cos x)^2\) \({}+\sin(5\pi+x)\times\cos(-x)\).

17.1. Mostre que \(A(x)=1-3\sin x\times\cos x\).

17.2. Determine o valor de \(A\left(\dfrac{13\pi}{4}\right)\).

17.3. Para um determinado valor de \(\alpha\in\left]-\dfrac{\pi}{2},\dfrac{\pi}{2}\right[\), sabe-se que \(\sin\alpha=-\dfrac34\). Determine o valor exato de \(A(\alpha)\).

17.1

(\sin x-\cos x)^2\) \(=\sin^2x-2\sin x\cos x\) \({}+\cos^2x\) \(=1-2\sin x\cos x

\sin(5\pi+x)\) \(=\sin(\pi+x)\) \(=-\sin x

\cos(-x)=\cos x

A(x)=1-2\sin x\cos x\) \({}-\sin x\cos x\) \(=1-3\sin x\cos x

17.2

\dfrac{13\pi}{4}=2\pi+\pi+\dfrac{\pi}{4}\) \(\Rightarrow\sin\dfrac{13\pi}{4}\) \(=-\dfrac{\sqrt2}{2},\ \cos\dfrac{13\pi}{4}\) \(=-\dfrac{\sqrt2}{2}

A\left(\dfrac{13\pi}{4}\right)\) \(=1-3\times\dfrac12\) \(=-\dfrac12

17.3 - α no 4.º quadrante

\cos^2\alpha\) \(=1-\dfrac{9}{16}=\dfrac{7}{16}\) \(\Rightarrow\cos\alpha=\dfrac{\sqrt7}{4}\quad\left(\alpha\in\left]-\dfrac{\pi}{2},0\right[\right)

A(\alpha)=1-3\times\left(-\dfrac34\right)\times\dfrac{\sqrt7}{4}\) \(=1+\dfrac{9\sqrt7}{16}

17.2. \(-\dfrac12\); 17.3. \(1+\dfrac{9\sqrt7}{16}\)

18. A expressão A(α)

Considere a expressão \(A(\alpha)=\sin(-11\pi+\alpha)\) \({}+\tan(-13\pi+\alpha)\) \({}\times\dfrac{2}{\cos\left(-\frac{3\pi}{2}-\alpha\right)}\), definida para \(\alpha\in\mathbb{R}\setminus\left\{\dfrac{k\pi}{2},\ k\in\mathbb{Z}\right\}\).

18.1. Mostre que, para todo o \(\alpha\) do domínio, se tem \(A(\alpha)=\dfrac{2}{\cos\alpha}-\sin\alpha\).

18.2. Determine o valor exato de \(A\left(\dfrac{4\pi}{3}\right)\).

18.1

\sin(-11\pi+\alpha)\) \(=\sin(\alpha-\pi)\) \(=-\sin\alpha

\tan(-13\pi+\alpha)=\tan\alpha

\cos\left(-\dfrac{3\pi}{2}-\alpha\right)\) \(=\cos\left(\dfrac{3\pi}{2}+\alpha\right)\) \(=\sin\alpha

A(\alpha)=-\sin\alpha\) \({}+\dfrac{\sin\alpha}{\cos\alpha}\times\dfrac{2}{\sin\alpha}\) \(=\dfrac{2}{\cos\alpha}-\sin\alpha

18.2

A\left(\dfrac{4\pi}{3}\right)=\dfrac{2}{-\frac12}-\left(-\dfrac{\sqrt3}{2}\right)\) \(=-4+\dfrac{\sqrt3}{2}

18.2. \(-4+\dfrac{\sqrt3}{2}\)

Além de \(k\pi\), como aparece \(\tan\alpha\) e a fração \(\frac{2}{\cos\alpha}\), é preciso excluir também \(\frac{\pi}{2}+k\pi\); daí o domínio \(\mathbb{R}\setminus\left\{\frac{k\pi}{2}\right\}\).

19. Tangente conhecida no 2.º quadrante

Considere \(\alpha\in\left]\dfrac{\pi}{2},\dfrac{3\pi}{2}\right[\) tal que \(\tan(-\alpha)=2\). Determine o valor exato de \(\cos(\alpha-\pi)-\cos\left(\dfrac{3\pi}{2}-\alpha\right)\).

Simplificar a expressão

\cos(\alpha-\pi)\) \({}-\cos\left(\dfrac{3\pi}{2}-\alpha\right)\) \(=-\cos\alpha-(-\sin\alpha)\) \(=\sin\alpha-\cos\alpha

Razões de α

\(\tan(-\alpha)=2\Leftrightarrow\tan\alpha=-2\). Em \(\left]\frac{\pi}{2},\frac{3\pi}{2}\right[\), a tangente só é negativa no 2.º quadrante.

1+4=\dfrac{1}{\cos^2\alpha}\) \(\Leftrightarrow\cos^2\alpha=\dfrac15\) \(\Rightarrow\cos\alpha=-\dfrac{\sqrt5}{5}

\sin\alpha=\tan\alpha\cos\alpha=\dfrac{2\sqrt5}{5}

\sin\alpha-\cos\alpha\) \(=\dfrac{2\sqrt5}{5}+\dfrac{\sqrt5}{5}\) \(=\dfrac{3\sqrt5}{5}

\(\dfrac{3\sqrt5}{5}\)

20. Cosseno conhecido no 4.º quadrante

Considere \(\alpha\in\left]\pi,2\pi\right[\) tal que \(4\cos(\alpha+\pi)=-1\). Determine o valor exato de \(\tan(\alpha-3\pi)-\sin(-\alpha-\pi)\).

4\cos(\alpha+\pi)=-1\) \(\Leftrightarrow-4\cos\alpha=-1\) \(\Leftrightarrow\cos\alpha=\dfrac14

Como \(\alpha\in\left]\pi,2\pi\right[\) e \(\cos\alpha\gt0\), \(\alpha\) está no 4.º quadrante:

\sin\alpha=-\sqrt{1-\dfrac{1}{16}}\) \(=-\dfrac{\sqrt{15}}{4}

\tan\alpha=\dfrac{-\frac{\sqrt{15}}{4}}{\frac14}=-\sqrt{15}

\tan(\alpha-3\pi)\) \({}-\sin(-\alpha-\pi)\) \(=\tan\alpha-(-\sin(\alpha+\pi))\) \(=\tan\alpha-\sin\alpha

-\sqrt{15}+\dfrac{\sqrt{15}}{4}=-\dfrac{3\sqrt{15}}{4}

\(-\dfrac{3\sqrt{15}}{4}\)

21. Seno de 3π/2 − α conhecido

Considere \(\alpha\in\left]-2\pi,-\pi\right[\), tal que \(\sin\left(\dfrac{3\pi}{2}-\alpha\right)=\dfrac58\). Determine o valor exato de \(\cos\left(-\alpha-\dfrac{3\pi}{2}\right)+\tan(-\alpha)\).

\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \(=-\cos\alpha=\dfrac58\) \(\Leftrightarrow\cos\alpha=-\dfrac58

\(\alpha\in\left]-2\pi,-\pi\right[\) corresponde a \(\left]0,\pi\right[\) (somando \(2\pi\)), onde o seno é positivo:

\sin\alpha=\sqrt{1-\dfrac{25}{64}}=\dfrac{\sqrt{39}}{8}

\tan\alpha=\dfrac{\frac{\sqrt{39}}{8}}{-\frac58}=-\dfrac{\sqrt{39}}{5}

\cos\left(-\alpha-\dfrac{3\pi}{2}\right)\) \(=\cos\left(\alpha+\dfrac{3\pi}{2}\right)\) \(=\sin\alpha

\tan(-\alpha)=-\tan\alpha

\sin\alpha-\tan\alpha\) \(=\dfrac{\sqrt{39}}{8}+\dfrac{\sqrt{39}}{5}\) \(=\dfrac{13\sqrt{39}}{40}

\(\dfrac{13\sqrt{39}}{40}\)

22. Dois ângulos relacionados

Relativamente a dois ângulos \(\alpha\) e \(\beta\), sabe-se que \(2\alpha-\beta=\dfrac{3\pi}{2}\) e \(\cos\alpha=-\dfrac45\).

Determine o valor exato de \(\sin(\alpha-\beta)\).

\beta=2\alpha-\dfrac{3\pi}{2}\) \(\Rightarrow\alpha-\beta=\alpha-2\alpha+\dfrac{3\pi}{2}\) \(=\dfrac{3\pi}{2}-\alpha

\sin(\alpha-\beta)\) \(=\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \(=-\cos\alpha=\dfrac45

\(\sin(\alpha-\beta)=\dfrac45\)

23. Cosseno de π/2 − α conhecido

Seja \(\alpha\in\left]\dfrac{\pi}{2},\dfrac{3\pi}{2}\right[\). Sabendo que \(\cos\left(\dfrac{\pi}{2}-\alpha\right)=-\dfrac15\), determine:

23.1. \(\cos\left(\dfrac{3\pi}{2}-\alpha\right)\) \({}+\cos\left(\dfrac{5\pi}{2}+\alpha\right)\) \({}-\sin(10\pi+\alpha)\)

23.2. \(\tan(\alpha-\pi)+\sin\left(\dfrac{\pi}{2}+\alpha\right)\)

Razões de α

\(\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha=-\frac15\). Em \(\left]\frac{\pi}{2},\frac{3\pi}{2}\right[\), o seno é negativo no 3.º quadrante:

\cos\alpha=-\sqrt{1-\dfrac{1}{25}}\) \(=-\dfrac{2\sqrt6}{5}

\tan\alpha=\dfrac{-\frac15}{-\frac{2\sqrt6}{5}}\) \(=\dfrac{1}{2\sqrt6}\) \(=\dfrac{\sqrt6}{12}

23.1

-\sin\alpha\) \({}+\cos\left(\dfrac{\pi}{2}+\alpha\right)-\sin\alpha\) \(=-\sin\alpha-\sin\alpha-\sin\alpha\) \(=-3\sin\alpha=\dfrac35

23.2

\tan\alpha+\cos\alpha\) \(=\dfrac{\sqrt6}{12}-\dfrac{2\sqrt6}{5}\) \(=\dfrac{5\sqrt6-24\sqrt6}{60}\) \(=-\dfrac{19\sqrt6}{60}

23.1. \(\dfrac35\); 23.2. \(-\dfrac{19\sqrt6}{60}\)

24. Valor numérico de uma expressão

Sendo \(\alpha\) a amplitude, em radianos, de um ângulo, calcule o valor numérico de \(\cos\left(\dfrac{5\pi}{2}+\alpha\right)-\tan(-\alpha)\), sabendo que \(\sin\left(\dfrac{3\pi}{2}-\alpha\right)=-\dfrac45\), com \(\alpha\in[0,\pi]\).

\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \(=-\cos\alpha=-\dfrac45\) \(\Leftrightarrow\cos\alpha=\dfrac45

Em \([0,\pi]\), com cosseno positivo, \(\alpha\) está no 1.º quadrante: \(\sin\alpha=\frac35\) e \(\tan\alpha=\frac34\).

\cos\left(\dfrac{5\pi}{2}+\alpha\right)\) \({}-\tan(-\alpha)\) \(=\cos\left(\dfrac{\pi}{2}+\alpha\right)+\tan\alpha\) \(=-\sin\alpha+\tan\alpha\) \(=-\dfrac35+\dfrac34=\dfrac{3}{20}

\(\dfrac{3}{20}\)

25. Seno de π − α conhecido

Sabendo que \(\alpha\in\left]\dfrac{\pi}{2},\dfrac{3\pi}{2}\right[\) e \(\sin(\pi-\alpha)=\dfrac35\), calcule o valor exato da expressão \(5\cos(\pi+\alpha)-4\tan(-\alpha)\).

\(\sin(\pi-\alpha)=\sin\alpha=\frac35\gt0\); em \(\left]\frac{\pi}{2},\frac{3\pi}{2}\right[\), \(\alpha\) está no 2.º quadrante.

\cos^2\alpha\) \(=1-\dfrac{9}{25}=\dfrac{16}{25}\) \(\Rightarrow\cos\alpha=-\dfrac45

\tan\alpha=\dfrac{\frac35}{-\frac45}=-\dfrac34

5\cos(\pi+\alpha)\) \({}-4\tan(-\alpha)\) \(=-5\cos\alpha+4\tan\alpha\) \(=4-3=1

\(1\)

26. Sem recorrer à calculadora

Determine o valor de \(1-\dfrac{\sqrt2}{\tan\alpha}\) sabendo que \(\alpha\in\left]\pi,\dfrac{3\pi}{2}\right[\) e que \(\sin\left(-\dfrac{\pi}{2}-\alpha\right)=\dfrac{\sqrt6}{3}\). Resolva esta questão sem recorrer à calculadora.

\sin\left(-\dfrac{\pi}{2}-\alpha\right)\) \(=-\sin\left(\dfrac{\pi}{2}+\alpha\right)\) \(=-\cos\alpha=\dfrac{\sqrt6}{3}\) \(\Leftrightarrow\cos\alpha=-\dfrac{\sqrt6}{3}

\tan^2\alpha+1\) \(=\dfrac{1}{\cos^2\alpha}=\dfrac{9}{6}\) \(\Leftrightarrow\tan^2\alpha=\dfrac12

Como \(\alpha\) está no 3.º quadrante, \(\tan\alpha\gt0\): \(\tan\alpha=\frac{\sqrt2}{2}\).

1-\dfrac{\sqrt2}{\frac{\sqrt2}{2}}=1-2=-1

\(-1\)

27. Ponto na reta x = 1

Na figura estão representados, em referencial o.n. \(Oxy\): a circunferência trigonométrica; a reta \(s\), de equação \(x=1\); o ponto \(A\left(1,-\sqrt8\right)\); o ângulo orientado, de amplitude \(\alpha\) radianos, cujos lados origem e extremidade são, respetivamente, o semieixo positivo \(Ox\) e a semirreta \(\dot{O}A\).

Circunferência trigonométrica, reta s de equação x igual a 1 e o ponto A de ordenada menos raiz de 8

Determine o valor exato de \(\tan(-\alpha)+\sin\left(-\dfrac{3\pi}{2}+\alpha\right)\).

A ordenada de \(A\) é a tangente de \(\alpha\): \(\tan\alpha=-\sqrt8\), com \(\alpha\) no 4.º quadrante.

1+8=\dfrac{1}{\cos^2\alpha}\) \(\Rightarrow\cos\alpha=\dfrac13\quad(4.^\text{o}\text{ Q})

\tan(-\alpha)\) \({}+\sin\left(\alpha-\dfrac{3\pi}{2}\right)\) \(=-\tan\alpha\) \({}+\sin\left(\alpha+\dfrac{\pi}{2}\right)\) \(=\sqrt8+\cos\alpha\) \(=2\sqrt2+\dfrac13

\(2\sqrt2+\dfrac13\)

28. Circunferência de raio 3

Na figura estão representados, num referencial o.n. \(Oxy\), a circunferência de centro \(O\) e raio \(3\) e o ângulo de amplitude \(\alpha\), que tem por lado origem o semieixo positivo \(Ox\) e por lado extremidade a semirreta \(\dot{O}A\).

Circunferência de centro O e raio 3 com o ponto A no terceiro quadrante

Sabe-se que a ordenada do ponto \(A\) é \(-\dfrac95\).

Determine o valor de \(4\cos\left(\dfrac{\pi}{2}+\alpha\right)\) \({}-\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \({}+\tan(2025\pi-\alpha)\).

Razões de α

3\sin\alpha=-\dfrac95\) \(\Leftrightarrow\sin\alpha=-\dfrac35

\cos\alpha=-\dfrac45\ (3.^\text{o}\text{ Q})

\tan\alpha=\dfrac34

Reduzir cada parcela

4\cos\left(\dfrac{\pi}{2}+\alpha\right)\) \(=-4\sin\alpha=\dfrac{12}{5}

-\sin\left(\dfrac{3\pi}{2}-\alpha\right)\) \(=\cos\alpha\) \(=-\dfrac45

\tan(2025\pi-\alpha)\) \(=\tan(-\alpha)\) \(=-\tan\alpha\) \(=-\dfrac34

\dfrac{12}{5}-\dfrac45-\dfrac34\) \(=\dfrac85-\dfrac34=\dfrac{17}{20}

\(\dfrac{17}{20}\)

Atenção ao coeficiente \(4\) da primeira parcela: quem o esquece obtém \(-\frac{19}{20}\); o valor correto é \(\frac{17}{20}\).

29. Qual é positivo? I

Seja \(\alpha\) um valor pertencente ao intervalo \(\left]-\dfrac{\pi}{2},0\right[\). Qual das expressões seguintes representa um número real positivo?

(A) \(\cos(\pi-\alpha)\)(B) \(\sin\left(-\dfrac{\pi}{2}+\alpha\right)\)(C) \(\cos\left(-\dfrac{\pi}{2}+\alpha\right)\)(D) \(\sin(10\pi-\alpha)\)

No 4.º quadrante, \(\sin\alpha\lt0\) e \(\cos\alpha\gt0\).

\text{(A)}\ -\cos\alpha\lt0

\text{(B)}\ -\cos\alpha\lt0

\text{(C)}\ \sin\alpha\lt0

\text{(D)}\ \sin(-\alpha)=-\sin\alpha\gt0

Opção (D).

30. Qual é positivo? II

Seja \(\alpha\) um valor pertencente ao intervalo \(\left]-\pi,0\right[\) tal que \(\tan(-\alpha)\times\cos\left(\alpha-\dfrac{\pi}{2}\right)\gt0\). Qual das expressões seguintes representa um número real positivo?

(A) \(\cos\alpha-\tan\alpha\)(B) \(\dfrac{\tan\alpha}{\sin\alpha}\)(C) \(\cos\alpha\times\sin\alpha\)(D) \(\sin\alpha+\cos\alpha\)

\tan(-\alpha)\times\cos\left(\alpha-\dfrac{\pi}{2}\right)\) \(=-\tan\alpha\times\sin\alpha\gt0\) \(\Leftrightarrow\tan\alpha\times\sin\alpha\lt0

Em \(\left]-\pi,0\right[\), \(\sin\alpha\lt0\); logo \(\tan\alpha\gt0\) e \(\alpha\) está no 3.º quadrante: \(\cos\alpha\lt0\).

(A) negativo menos positivo; (B) positivo sobre negativo; (D) soma de negativos. (C) é o produto de dois negativos, logo positivo.

Opção (C).

31. Qual é o valor de sin α?

Seja \(\alpha\) a amplitude de um ângulo orientado tal que \(\sin\left(-\dfrac{\pi}{2}+\alpha\right)\gt0\) e \(\tan(\pi+\alpha)\lt0\). Qual das expressões seguintes dá o valor de \(\sin\alpha\)?

(A) \(\sqrt{1-\cos^2\alpha}\)(B) \(\sqrt{1+\cos^2\alpha}\)(C) \(-\sqrt{1-\cos^2\alpha}\)(D) \(-\sqrt{1+\cos^2\alpha}\)

\sin\left(\alpha-\dfrac{\pi}{2}\right)\) \(=-\cos\alpha\gt0\) \(\Leftrightarrow\cos\alpha\lt0

\tan(\pi+\alpha)=\tan\alpha\lt0

Cosseno negativo e tangente negativa: \(\alpha\) está no 2.º quadrante, logo \(\sin\alpha\gt0\). Da fórmula fundamental, \(\sin\alpha=\sqrt{1-\cos^2\alpha}\).

Opção (A).

32. Qual é a afirmação verdadeira?

Seja \(\alpha\in\left[0,2\pi\right[\) tal que \(\cos(\pi+\alpha)\gt0\) e \(\cos\left(\dfrac{\pi}{2}+\alpha\right)\gt0\). Qual das seguintes afirmações é verdadeira?

(A) \(\sin(\pi+\alpha)\lt0\)(B) \(\tan(\alpha-\pi)\lt0\)(C) \(\tan(\pi+\alpha)\gt0\)(D) \(\sin\left(\dfrac{\pi}{2}+\alpha\right)\gt0\)

\cos(\pi+\alpha)\) \(=-\cos\alpha\gt0\) \(\Leftrightarrow\cos\alpha\lt0

\cos\left(\dfrac{\pi}{2}+\alpha\right)\) \(=-\sin\alpha\gt0\) \(\Leftrightarrow\sin\alpha\lt0

\(\alpha\) está no 3.º quadrante, logo \(\tan\alpha\gt0\).

(A) \(\sin(\pi+\alpha)=-\sin\alpha\gt0\): falsa. (B) \(\tan(\alpha-\pi)=\tan\alpha\gt0\): falsa. (C) \(\tan(\pi+\alpha)=\tan\alpha\gt0\): verdadeira. (D) \(\sin\left(\frac{\pi}{2}+\alpha\right)=\cos\alpha\lt0\): falsa.

Opção (C).

33. Ângulos de um trapézio retângulo

Na figura está representado um trapézio retângulo, no qual estão assinalados os ângulos \(\alpha\) e \(\beta\).

Trapézio retângulo com o ângulo beta no vértice superior direito e alfa no inferior direito

Qual das condições seguintes é verdadeira, para qualquer trapézio retângulo?

(A) \(\sin\alpha=\cos\beta\)(B) \(\sin\alpha=-\cos\beta\)(C) \(\sin\alpha=\sin\beta\)(D) \(\sin\alpha=-\sin\beta\)

As bases do trapézio são paralelas, logo os ângulos \(\alpha\) e \(\beta\) (adjacentes ao mesmo lado oblíquo) são suplementares: \(\beta=\pi-\alpha\).

\sin\beta=\sin(\pi-\alpha)=\sin\alpha

Opção (C).