Question

compute sin 15degree from the functions of 60degree and 45degree​

Answer 1

Explanation is written in the image linked above.

Answer 2

Step-by-step explanation:

$\sin(45) = \frac{1}{ \sqrt{2} } \\ \sin(60) = \frac{ \sqrt{3} }{2} \\ \cos(45) = \frac{1}{ \sqrt{2} } \\ \cos(60) = \frac{1}{2} \\ \sin( \alpha - \beta ) = \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta ) \\ \\ \sin(60 - 45) = \sin(60) \cos(45) + \cos(60) \sin(45) \\ = > \sin(15) = ( \frac{ \sqrt{3} }{2} )( \frac{1}{ \sqrt{2} } ) + ( \frac{1}{2} )( \frac{1}{ \sqrt{2} } ) \\ = > \sin(15) = \frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } \\ = > \sin(15) = \frac{ \sqrt{3} + 1 }{2 \sqrt{2} }$