Question

Sin(60-theta)sin(60+theta)​

Answer 1

Explanation:

`\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ ( √(3) )/(2) \cos( \theta) - (1)/(2) \sin( \theta) \} \{ ( √(3) )/(2) \cos( \theta) + (1)/(2) \sin( \theta) \} \\ \\`

from difference of two squares:

`{ \boxed{(a - b)(a + b) = ( {a}^(2) - {b}^(2) ) }}`

therefore:

`= \{ {( ( √(3) )/(2)) }^(2) { \cos }^(2) \theta \} - \{ {( ( √(3) )/(2) )}^(2) { \sin }^(2) \theta \} \\ \\ = (3)/(4) { \cos }^(2) \theta - (3)/(4) { \sin}^(2) \theta`

factorise out ¾ :

`= (3)/(4) ( { \cos }^(2) \theta - { \sin}^(2) \theta) \\ \\ = { \boxed{ (3)/(4) \cos(2 \theta) }}`