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Question Number 35877 by 0123456789 last updated on 25/May/18
If cos A=(3/4) , then 32 sin ((A/2)) sin (((5A)/2))=
$$\mathrm{If}\:\:\mathrm{cos}\:{A}=\frac{\mathrm{3}}{\mathrm{4}}\:,\:\mathrm{then}\:\mathrm{32}\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{5}{A}}{\mathrm{2}}\right)= \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18
32sin((A/2))sin(((5A)/2)) =16×(cos2A−cos3A) =16×{(2cos^2 A−1)−(4cos^3 A−3cosA)} =16×{(((18)/(16))−1)−(4×((27)/(64))−(9/4))} =16×{((1/8))−(((27)/(16))−(9/4))} =16×{(1/8)−(((27−36)/(16)))} =16×{((2+9)/(16))} =11
$$\mathrm{32}{sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{5}{A}}{\mathrm{2}}\right) \\ $$$$=\mathrm{16}×\left({cos}\mathrm{2}{A}−{cos}\mathrm{3}{A}\right) \\ $$$$=\mathrm{16}×\left\{\left(\mathrm{2}{cos}^{\mathrm{2}} {A}−\mathrm{1}\right)−\left(\mathrm{4}{cos}^{\mathrm{3}} {A}−\mathrm{3}{cosA}\right)\right\} \\ $$$$=\mathrm{16}×\left\{\left(\frac{\mathrm{18}}{\mathrm{16}}−\mathrm{1}\right)−\left(\mathrm{4}×\frac{\mathrm{27}}{\mathrm{64}}−\frac{\mathrm{9}}{\mathrm{4}}\right)\right\} \\ $$$$=\mathrm{16}×\left\{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)−\left(\frac{\mathrm{27}}{\mathrm{16}}−\frac{\mathrm{9}}{\mathrm{4}}\right)\right\} \\ $$$$=\mathrm{16}×\left\{\frac{\mathrm{1}}{\mathrm{8}}−\left(\frac{\mathrm{27}−\mathrm{36}}{\mathrm{16}}\right)\right\} \\ $$$$=\mathrm{16}×\left\{\frac{\mathrm{2}+\mathrm{9}}{\mathrm{16}}\right\} \\ $$$$=\mathrm{11} \\ $$

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