Ensino Médio ⇒ Transformações Trigonométricas Tópico resolvido

[mahriana]

Calcule [tex3]\mathbf{sen^2\frac{\pi }{12}-cos^2\frac{\pi }{12}+tg\frac{\pi }{3}+tg\frac{14\pi }{3}}[/tex3]

Resposta

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[tex3]\frac{-\sqrt{3}}{2}[/tex3]

[Vinisth]

Olá mahriana ,

Lembre-se que [tex3]cos(x)\pm sen(x)=\pm \sqrt{2}.sen(\frac{\pi}{4}\pm x)[/tex3]

[tex3]E = \mathbf{sen^2\frac{\pi }{12}-cos^2\frac{\pi }{12}+tg\frac{\pi }{3}+tg\frac{14\pi }{3}}[/tex3]
[tex3]E = (sen\frac{\pi }{12}+cos\frac{\pi }{12})(sen\frac{\pi }{12}-cos\frac{\pi }{12})+tg\frac{\pi }{3}+tg\frac{14\pi }{3}[/tex3]
[tex3]E = \sqrt{2}.sen\left(\frac{\pi}{4}+\frac{\pi}{12}\right).(-\sqrt{2})sen\left(\frac{\pi}{4}-\frac{\pi}{12}\right)+tg\frac{\pi }{3}-tg\frac{\pi }{3}[/tex3]
[tex3]E = -2.\frac{\sqrt{3}}{2}.\frac{1}{2}=\boxed{\frac{-\sqrt{3}}{2}}[/tex3]

Abraço.

PS: Editei aqui. Errei no final.

[jhonim]

Olá mahriana,

Simplificando:

[tex3]\mathbf{sen^2\frac{\pi }{12}-cos^2\frac{\pi }{12}+tg\frac{\pi }{3}+tg\frac{14\pi }{3}}=-cos\left(2\cdot\frac{\pi}{12}\right)+tg\left(\frac{\pi}{3}\right)+tg\left(4\pi+\frac{2\pi}{3}\right)\\\\\\\mathbf{sen^2\frac{\pi }{12}-cos^2\frac{\pi }{12}+tg\frac{\pi }{3}+tg\frac{14\pi }{3}}=-cos\left(\frac{\pi}{6}\right)+tg\left(\frac{\pi}{3}\right)-tg\left(\frac{\pi}{3}\right)=-cos\left(\frac{\pi}{6}\right)\\\\\\\mathbf{sen^2\frac{\pi }{12}-cos^2\frac{\pi }{12}+tg\frac{\pi }{3}+tg\frac{14\pi }{3}}=-cos\left(\frac{\pi}{6}\right)=-cos(30)=\frac{-\sqrt{3}}{2}[/tex3]


Até mais e bons estudos.