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Question Number 133929 by bemath last updated on 25/Feb/21

$$\cos {}^{2}76°+\cos {}^{2}16°-\cos 76°.\cos 16°=?$$

Answered by liberty last updated on 25/Feb/21

$$\Rightarrow a^2+b^2={(a+b)}^2-2ab$$$$$$$$\cos {}^{2}76°+\cos {}^{2}16°-\cos 76°.\cos 16°=$$$${[\cos 76°+\cos 16°]}^2-3\cos 76°\cos 16°$$$$={\left[2\cos 46°\cos 30°\right]}^2-\frac{3}{2}[\cos 92°+\cos 60°]$$$$=3\cos {}^{2}46°-\frac{3}{2}\cos 92°-\frac{3}{4}$$$$=3[\frac{1}{2}+\frac{1}{2}\cos 92°]-\frac{3}{2}\cos 92°-\frac{3}{4}$$$$=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}$$

Answered by EDWIN88 last updated on 25/Feb/21

$$\cos {}^{2}76°+\cos {}^{2}16°-\cos 76°.\cos 16°=$$$$\frac{1}{2}[1+\cos 152°+1+\cos 32°-\cos 92°-\cos 60°]$$$$=\frac{1}{2}[2-\frac{1}{2}+\cos 152°+\cos 32°-\cos 92°]$$$$=\frac{1}{2}[\frac{3}{2}+2\cos 92°\cos 60°-\cos 92°]$$$$=\frac{1}{2}[\frac{3}{2}+\cos 92°-\cos 92°]=\frac{3}{4}$$$$$$

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