Question

If `cos "A" = 3/4,`then 32 sin`"A"/2 cos  (5"A")/2` = ?

विकल्प

• `sqrt11`

• `- sqrt7`

• 16

• - 32

Answer 1

`- sqrt7`

Explanation:

`cos "A" = 3/4 => sin "A" = sqrt7/4`

`32 sin  "A"/2 cos  "5A"/2 = 16 (sin 3"A" - sin "2A")`

= 16(3 sin A - 4 sin³ A - 2 sin A cos A)

`= 16[3 (sqrt7)/4 - 4((7sqrt7)/64) - 2(sqrt7/4)(3/4)]`

= - `sqrt7`