1. Simplificação de expressões
1
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) \(\cos\left(\dfrac{5\pi}{2}+\alpha\right)-\sin\left(\alpha-\dfrac{7\pi}{2}\right)-\cos(7\pi-\alpha)+2\sin(\alpha+4\pi)\)
a)
\(\sin(\pi-\alpha)-\sin(2\pi-\alpha)+\cos(-\alpha)+\cos(\pi-\alpha)=\)
\(=\sin\alpha+\sin\alpha+\cos\alpha-\cos\alpha=2\sin\alpha\)
b)
\(\tan(\pi-\alpha)-\tan(2\pi-\alpha)-3\sin(-\alpha)-2\sin(2\pi+\alpha)=\)
\(=-\tan\alpha+\tan\alpha+3\sin\alpha-2\sin\alpha=\sin\alpha\)
c)
\(\sin(-\alpha)\sin(\pi+\alpha)-\cos(-\alpha)\cos(\pi-\alpha)=\)
\(=\sin\alpha\times(-\sin\alpha)-\cos\alpha\times(-\cos\alpha)=\sin^2\alpha+\cos^2\alpha=1\)
d)
\(5\sin\left(\dfrac{\pi}{2}-\alpha\right)+\cos\left(\dfrac{\pi}{2}+\alpha\right)-\sin(\pi+\alpha)+3\cos(\pi+\alpha)=\)
\(=5\cos\alpha-\sin\alpha+\sin\alpha-3\cos\alpha=2\cos\alpha\)
e)
\(\sin\left(\dfrac{3\pi}{2}-\alpha\right)+\cos(2\pi-\alpha)-\tan(-\alpha)\cos(2\pi-\alpha)+\cos\left(\dfrac{3\pi}{2}+\alpha\right)=\)
\(=-\cos\alpha+\cos\alpha+\tan\alpha\times\cos\alpha+\sin\alpha=\dfrac{\sin\alpha}{\cos\alpha}\times\cos\alpha+\sin\alpha=2\sin\alpha\)
f)
\(\sin(\alpha-\pi)-\tan(\alpha+\pi)-\tan(2\pi-\alpha)-5\cos\left(\alpha-\dfrac{3\pi}{2}\right)=\)
\(=-\sin\alpha-\tan\alpha+\tan\alpha+5\sin\alpha=4\sin\alpha\)
g)
\(\sin(\alpha-5\pi)\times\tan(-\alpha-4\pi)-3\sin\left(\alpha-\dfrac{\pi}{2}\right)+2\cos(7\pi-\alpha)=\)
\(=-\sin\alpha\times(-\tan\alpha)+3\cos\alpha-2\cos\alpha=\sin\alpha\times\dfrac{\sin\alpha}{\cos\alpha}+\cos\alpha=\)
\(=\dfrac{\sin^2\alpha}{\cos\alpha}+\cos\alpha=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha}=\dfrac{1}{\cos\alpha}\)
h)
\(\cos\left(\dfrac{5\pi}{2}+\alpha\right)-\sin\left(\alpha-\dfrac{7\pi}{2}\right)-\cos(7\pi-\alpha)+2\sin(\alpha+4\pi)=\)
\(=-\sin\alpha-\cos\alpha+\cos\alpha+2\sin\alpha=\sin\alpha\)
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) \(\sin\alpha\).
e) \(2\sin\alpha\); f) \(4\sin\alpha\); g) \(\dfrac{1}{\cos\alpha}\); h) \(\sin\alpha\).
2. Valor lógico de uma proposição
2
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 \(\sin x>0\), então a tangente 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:
\(-\cos x<0\Leftrightarrow\cos(-\pi-x)<0\)
Portanto, a proposição é verdadeira.
3. Demonstração de identidades
3
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) \(2\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}+x\right)-\cos(2\pi)=\left[1-\sqrt{2}\,\sin(2\pi+x)\right]\left[1-\sqrt{2}\,\sin(\pi+x)\right]\)
a)
\(1-\dfrac{\cos(2\pi+x)}{\tan(2\pi+x)\sin x+\cos(2\pi-x)}=1-\dfrac{\cos x}{\tan x\times\sin x+\cos x}=\)
\(=1-\dfrac{\cos x}{\dfrac{\sin x}{\cos x}\times\sin x+\cos x}=1-\dfrac{\cos x}{\dfrac{\sin^2 x+\cos^2 x}{\cos x}}=1-\dfrac{\cos x}{\dfrac{1}{\cos x}}=1-\cos^2 x=\sin^2 x\)
b)
\(2\sin\left(\dfrac{\pi}{2}-x\right)\sin\left(\dfrac{\pi}{2}+x\right)-\cos(2\pi)=2\cos x\cos x-\cos(2\pi)=2\cos^2 x-1=\)
\(=2(1-\sin^2 x)-1=2-2\sin^2 x-1=1-2\sin^2 x=1-\left(\sqrt{2}\,\sin x\right)^2=\)
\(=\left(1-\sqrt{2}\,\sin x\right)\left(1+\sqrt{2}\,\sin x\right)=\left(1-\sqrt{2}\,\sin(2\pi+x)\right)\left(1-\sqrt{2}\,\sin(\pi+x)\right)\)
Em ambos os casos as identidades ficam demonstradas. \(\blacksquare\)
4. Valor de expressões
4
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) \(\dfrac{\sin\left(-\dfrac{17\pi}{3}\right)}{\cos\dfrac{17\pi}{4}}+\sqrt{6}\,\sin\dfrac{37\pi}{6}\)
a)
\(\sin\left(-\dfrac{\pi}{6}\right)+\sqrt{6}\,\sin\left(-\dfrac{\pi}{4}\right)\sin\left(-\dfrac{\pi}{3}\right)=-\sin\dfrac{\pi}{6}+\left(-\sqrt{6}\,\sin\dfrac{\pi}{4}\right)\left(-\sin\dfrac{\pi}{3}\right)=\)
\(=-\sin\dfrac{\pi}{6}+\sqrt{6}\,\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{3}=-\dfrac{1}{2}+\sqrt{6}\times\dfrac{\sqrt{2}}{2}\times\dfrac{\sqrt{3}}{2}=-\dfrac{1}{2}+\dfrac{6}{4}=-\dfrac{1}{2}+\dfrac{3}{2}=1\)
b)
\(\sin\left(-\dfrac{3\pi}{4}\right)-\cos\left(-\dfrac{\pi}{4}\right)+\sqrt{5-\sqrt{3}\,\tan\left(-\dfrac{\pi}{3}\right)}=-\sin\dfrac{\pi}{4}-\cos\dfrac{\pi}{4}+\sqrt{5+\sqrt{3}\,\tan\dfrac{\pi}{3}}=\)
\(=-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}+\sqrt{5+\sqrt{3}\times\sqrt{3}}=-\sqrt{2}+\sqrt{8}=-\sqrt{2}+2\sqrt{2}=\sqrt{2}\)
c)
\(\dfrac{\sin\dfrac{5\pi}{6}+\cos\left(-\dfrac{3\pi}{4}\right)}{2+\cos(-\pi)+\cos\left(-\dfrac{2\pi}{3}\right)-\sin\left(-\dfrac{3\pi}{4}\right)}=\dfrac{\sin\dfrac{\pi}{6}-\cos\dfrac{\pi}{4}}{2-1-\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{4}}=\dfrac{\dfrac{1}{2}-\dfrac{\sqrt{2}}{2}}{1-\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}}=\)
\(=\dfrac{\dfrac{1-\sqrt{2}}{2}}{\dfrac{1+\sqrt{2}}{2}}=\dfrac{1-\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(1-\sqrt{2}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{1-2\sqrt{2}+2}{1-2}=\dfrac{3-2\sqrt{2}}{-1}=2\sqrt{2}-3\)
d)
\(\dfrac{\sin\left(-\dfrac{17\pi}{3}\right)}{\cos\dfrac{17\pi}{4}}+\sqrt{6}\,\sin\dfrac{37\pi}{6}=\dfrac{\sin\left(-\dfrac{17\pi}{3}+6\pi\right)}{\cos\left(\dfrac{17\pi}{4}-4\pi\right)}+\sqrt{6}\,\sin\left(\dfrac{37\pi}{6}-6\pi\right)=\)
\(=\dfrac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{4}}+\sqrt{6}\,\sin\dfrac{\pi}{6}=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{\sqrt{2}}{2}}+\sqrt{6}\times\dfrac{1}{2}=\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{6}}{2}=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{2}=\sqrt{6}\)
a) \(1\); b) \(\sqrt{2}\); c) \(2\sqrt{2}-3\); d) \(\sqrt{6}\).
5. Determinação de x e cálculo
5
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 \(-1\).