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Question Number 149259 by john_santu last updated on 04/Aug/21

cos^(−1) ((√(sin (41°)sin (19°)+(3/4))) )=?

$$\:\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{sin}\:\left(\mathrm{41}°\right)\mathrm{sin}\:\left(\mathrm{19}°\right)+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)=? \\ $$

Answered by john_santu last updated on 04/Aug/21

⇔sin 41°sin 19°=((−2sin 4l°sin 19°)/(−2)) =((cos 60°−cos 22°)/(−2))=((cos 22°−(1/2))/2) =(1/2)cos 22°−(1/4) ⇒cos^(−1) ((√((1/2)cos 22°+(1/2))) )= cos^(−1) ((√((1/2)(2cos^2 11°−1+1)))) cos^(−1) ((√(cos^2 11°)))=cos^(−1) (cos 11°) = 11°

$$\Leftrightarrow\mathrm{sin}\:\mathrm{41}°\mathrm{sin}\:\mathrm{19}°=\frac{−\mathrm{2sin}\:\mathrm{4l}°\mathrm{sin}\:\mathrm{19}°}{−\mathrm{2}} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{60}°−\mathrm{cos}\:\mathrm{22}°}{−\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{22}°−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{22}°+\frac{\mathrm{1}}{\mathrm{2}}}\:\right)= \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{11}°−\mathrm{1}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:^{\mathrm{2}} \mathrm{11}°}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{11}°\right) \\ $$$$=\:\mathrm{11}°\: \\ $$

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