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Question Number 146436 by mathdanisur last updated on 13/Jul/21

((sin^2 25 - sin^2 5)/(sin 20)) = ?

$$\frac{{sin}^{\mathrm{2}} \:\mathrm{25}\:-\:{sin}^{\mathrm{2}} \:\mathrm{5}}{{sin}\:\mathrm{20}}\:=\:? \\ $$

Answered by gsk2684 last updated on 13/Jul/21

((sin (25^0 +5^0 ) sin (25^0 −5^0 ))/(sin 20^0 )) =sin 30^0 =(1/2)

$$\frac{\mathrm{sin}\:\left(\mathrm{25}^{\mathrm{0}} +\mathrm{5}^{\mathrm{0}} \right)\:\mathrm{sin}\:\left(\mathrm{25}^{\mathrm{0}} −\mathrm{5}^{\mathrm{0}} \right)}{\mathrm{sin}\:\mathrm{20}^{\mathrm{0}} } \\ $$$$=\mathrm{sin}\:\mathrm{30}^{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 13/Jul/21

thanks Ser

$${thanks}\:{Ser} \\ $$

Answered by liberty last updated on 13/Jul/21

((sin^2 25°−sin^2 5°)/(sin 20°)) = (((sin 25°+sin 5°)(sin 25°−sin 5°))/(sin 20°)) = (((2sin 15° cos 10°)(2cos 15°sin 10°))/(sin 20°)) = sin 30°=(1/2)

$$\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{25}°−\mathrm{sin}\:^{\mathrm{2}} \mathrm{5}°}{\mathrm{sin}\:\mathrm{20}°}\:=\:\frac{\left(\mathrm{sin}\:\mathrm{25}°+\mathrm{sin}\:\mathrm{5}°\right)\left(\mathrm{sin}\:\mathrm{25}°−\mathrm{sin}\:\mathrm{5}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$=\:\frac{\left(\mathrm{2sin}\:\mathrm{15}°\:\cancel{\mathrm{cos}\:\mathrm{10}}°\right)\left(\cancel{\mathrm{2}cos}\:\mathrm{15}°\cancel{\mathrm{sin}\:}\mathrm{1}\cancel{\mathrm{0}°}\right)}{\cancel{\mathrm{sin}\:\mathrm{20}°}} \\ $$$$=\:\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 13/Jul/21

thanks Ser

$${thanks}\:{Ser} \\ $$

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