Question

\(\bf sin15\)

\(\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
\(\dfrac{\sqrt{6}-\sqrt{2}}{4}\)
\(-\dfrac{\sqrt{6}-\sqrt{2}}{4}\)
\(-\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

Answer 1

@jim_thompson5910

Answer 2

15 = 45 - 30

use

sin(x-y) = sin(x)cos(y) - cos(x)sin(y)

Answer 3

x = 45
y = 30

Answer 4

I am thinking the second one

Answer 5

yeah I'm getting

\[\Large \frac{\sqrt{6}-\sqrt{2}}{4}\]

as well