Question

Can you solve this for me?
sin((pi/6)+x)=cos((pi/4)-x)

Answer 1

\[\sin(x)=\cos(\frac{\pi}{2}-x) ; \cos(x)=\sin(\frac{\pi}{2}-x)\]
\[\sin(\frac{\pi}{6}+x)=\sin(\frac{\pi}{2}-(\frac{\pi}{4}-x))\]
\[\cos(\frac{\pi}{4}-x)=\cos(\frac{\pi}{2}-(\frac{\pi}{6}+x))\]

Answer 2

i used those two top equations to rewrite what you have

Answer 3

no I suppose we should put something in there like 2npi

Answer 4

\[\sin(\frac{\pi}{6}+x)=\sin(\frac{\pi}{2}-(\frac{\pi}{4}-x)+2n \pi)\]
\[\cos(\frac{\pi}{4}-x)=\cos(\frac{\pi}{2}-(\frac{\pi}{6}+x)+2n \pi)\]

Answer 5

where n is an integer

Answer 6

now to solve both equations set the insides equal and solve for x

Answer 7

Thanks a lot:)

Answer 8

:)

Answer 9

\[\sin(\frac{\pi}{6}+x)=\cos(\frac{\pi}{4}-x)\]

Answer 10

\[\sin\frac{\pi}{6}cosx+\cos\frac{\pi}{6}sinx=\cos\frac{\pi}{4}cosx+\sin\frac{\pi}{4}sinx\]

Answer 11

\[\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx\]

Answer 12

\[(\frac{\sqrt{3}-\sqrt{2}}{2})sinx=(\frac{\sqrt{2}-1}{2})cosx\]

Answer 13

\[sinx=(\frac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}})cosx\]

Answer 14

\[tanx=\frac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\]

Answer 15

\[x=\frac{7\pi}{24}\]