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Question Number 149141 by mathdanisur last updated on 03/Aug/21

((cos^2 (70) - sin^2 (70))/(sin(100) + sin(20))) = ?

$$\frac{{cos}^{\mathrm{2}} \left(\mathrm{70}\right)\:-\:{sin}^{\mathrm{2}} \left(\mathrm{70}\right)}{{sin}\left(\mathrm{100}\right)\:+\:{sin}\left(\mathrm{20}\right)}\:=\:? \\ $$

Commented by liberty last updated on 03/Aug/21

=(((cos 70°−sin 70°)(cos 70°+sin 70°))/(sin 100°+sin 20°)) =(((sin 20°−sin 70°)(sin 20°+sin 70°))/(2sin 60°cos 40°)) =((2cos 45°sin (−25°)(2sin 45°cos 25)/( (√3) sin 50°)) =((−sin 50°)/( (√3)sin 50°))=−(1/3)(√3)

$$=\frac{\left(\mathrm{cos}\:\mathrm{70}°−\mathrm{sin}\:\mathrm{70}°\right)\left(\mathrm{cos}\:\mathrm{70}°+\mathrm{sin}\:\mathrm{70}°\right)}{\mathrm{sin}\:\mathrm{100}°+\mathrm{sin}\:\mathrm{20}°} \\ $$$$=\frac{\left(\mathrm{sin}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{70}°\right)\left(\mathrm{sin}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{70}°\right)}{\mathrm{2sin}\:\mathrm{60}°\mathrm{cos}\:\mathrm{40}°} \\ $$$$=\frac{\mathrm{2cos}\:\mathrm{45}°\mathrm{sin}\:\left(−\mathrm{25}°\right)\left(\mathrm{2sin}\:\mathrm{45}°\mathrm{cos}\:\mathrm{25}\right.}{\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{50}°} \\ $$$$=\frac{−\mathrm{sin}\:\mathrm{50}°}{\:\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{50}°}=−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{3}} \\ $$

Commented by mathdanisur last updated on 03/Aug/21

Ser, Thank You

$${Ser},\:{Thank}\:{You} \\ $$

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