Question

Trigonometrie de clasa a noua.
Demonstrati ca sunt egale:
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Answer 1

$\frac{1+sin^2\frac{5\pi}{12}}{1+cos^2\frac{5\pi}{12}}=\frac{6+\sqrt{3}}{6-\sqrt{3}}\\ \\ \Leftrightarrow (6-\sqrt{3})(1+sin^2\frac{5\pi}{12})=(6+\sqrt{3})(1+cos^2\frac{5\pi}{12})$

$\Leftrightarrow 6+6sin^2\frac{5\pi}{12}-\sqrt{3}-\sqrt{3}sin^2\frac{5\pi}{12}=6+6cos^2\frac{5\pi}{12}+\sqrt{3}+\sqrt{3}cos^2\frac{5\pi}{12} \ |-6$

$\Leftrightarrow 6sin^2\frac{5\pi}{12}-6cos^2\frac{5\pi}{12}=2\sqrt{3}+\sqrt{3}(sin^2\frac{5\pi}{12}+cos^2\frac{5\pi}{12})$

$\Leftrightarrow 6(sin^2\frac{5\pi}{12}-cos^2\frac{5\pi}{12})=2\sqrt{3}+\sqrt{3}\cdot 1 \ |:6$

$\Leftrightarrow sin^2\frac{5\pi}{12}-cos^2\frac{5\pi}{12}=\frac{3\sqrt{3}}{6} \ |\cdot (-1)$

$\Leftrightarrow cos^2\frac{5\pi}{12}-sin^2\frac{5\pi}{12}=-\frac{\sqrt{3}}{2}$

$\Leftrightarrow cos(2\cdot \frac{5\pi}{12})=-\frac{\sqrt{3}}{2}$

$\Leftrightarrow cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$

$\frac{\pi}{2}< \frac{5\pi}{6}< \pi \Rightarrow \frac{5\pi}{6} \in \ cadranul \ II \Rightarrow cos \frac{5\pi}{6}=-cos(\pi-\frac{5\pi}{6})=-cos\frac{\pi}{6}$

$\Leftrightarrow -cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}, \ adevarat$

$Am \ folosit: \\ \\ \boxed{sin^2x+cos^2x=1}\\ \\ \boxed{cos^2x-sin^2x=cos2x}$

Answer 2

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