Question

Sin(60-theta)sin(60+theta)​

Answer 1

Answer:

It is a double angle of cosine

Answer 2

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \}$

from difference of two squares:

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

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$