If cos⁡A=34 then 32sin⁡A2⋅sin⁡5A2=

32sin⁡A2⋅sin⁡5A2=162sin5A2sinA2 =16cos5A2-A2-cos5A2+A2 =16cos2A-cos3A =162cos2A-1-4cos3A-3cosA =162×916-1-1642764-334 =18-16-27-36=2+9=11

If cos⁡A=34 then 32sin⁡A2⋅sin⁡5A2=

32sin⁡A2⋅sin⁡5A2=162sin5A2sinA2 =16cos5A2-A2-cos5A2+A2 =16cos2A-cos3A =162cos2A-1-4cos3A-3cosA =162×916-1-1642764-334 =18-16-27-36=2+9=11

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$If \cos A = \frac{3}{4} then 32 \sin \frac{A}{2} \cdot \sin \frac{5 A}{2} =$

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$11$

-10

answer is C.

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Detailed Solution

$32 \sin \frac{A}{2} \cdot \sin \frac{5 A}{2} = 16 (2 \sin \frac{5 A}{2} \sin \frac{A}{2}) = 16 (\cos (\frac{5 A}{2} - \frac{A}{2}) - \cos (\frac{5 A}{2} + \frac{A}{2})) = 16 (\cos 2 A - \cos 3 A) = 16 ((2 \cos^{2} A - 1) - (4 \cos^{3} A - 3 \cos A)) = 16 (2 \times \frac{9}{16} - 1) - 16 (4 (\frac{27}{64}) - 3 (\frac{3}{4})) = (18 - 16) - (27 - 36) = 2 + 9 = 11$