Question

prove.
sin [(pi over 4)+x] + sin [(pi over 4)-x] = sqrt{2} cosx

Answer 1

i think expanding both of those things might help

Answer 2

use ... sin(a=b)= sin(a)cos(b) +cos(a)sin(b)
and sin(a-b)= sin(a)cos(b) -cos(a)sin(b)

Answer 3

all i know is that i will have to use addition indentities.

Answer 4

\[[\sin(\frac{\pi}{4})\cos(x)+\sin(x)\cos(\frac{\pi}{4})]+[\sin(\frac{\pi}{4})\cos(x)-\sin(x)\cos(\frac{\pi}{4})]\]

Answer 5

\[\frac{\sqrt{2}}{2} \cos(x)+\sin(x) \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(x)-\sin(x)\frac{\sqrt{2}}{2}\]

Answer 6

combine like terms and you will be done

Answer 7

its 2sin(a)cos(b)....now plug valus.....(2/\[\sqrt{2}\] ) sinx

Answer 8

\[\sqrt{2}\] sin x

Answer 9

you rock myininaya!!!
thank you so much!!!

and i appreciate your work and help Sam_unleashed!
:)