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Hannah_Ahn
prove. sin [(pi over 4)+x] + sin [(pi over 4)-x] = sqrt{2} cosx
i think expanding both of those things might help
use ... sin(a=b)= sin(a)cos(b) +cos(a)sin(b) and sin(a-b)= sin(a)cos(b) -cos(a)sin(b)
all i know is that i will have to use addition indentities.
\[[\sin(\frac{\pi}{4})\cos(x)+\sin(x)\cos(\frac{\pi}{4})]+[\sin(\frac{\pi}{4})\cos(x)-\sin(x)\cos(\frac{\pi}{4})]\]
\[\frac{\sqrt{2}}{2} \cos(x)+\sin(x) \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(x)-\sin(x)\frac{\sqrt{2}}{2}\]
combine like terms and you will be done
its 2sin(a)cos(b)....now plug valus.....(2/\[\sqrt{2}\] ) sinx
\[\sqrt{2}\] sin x
you rock myininaya!!! thank you so much!!! and i appreciate your work and help Sam_unleashed! :)