Question

Rewrite in terms of sin(x) and cos(x).
sin(x+7π/4)

Answer 1

\[A\cos(x)+B\sin(x)=\sqrt{A^2+B^2}\sin(x+\theta)\] you have to go from the right to the left, which is kind of odd but doable

Answer 2

So I plug what I know into the right side of that equation?

Answer 3

you need A and B

Answer 4

\[\sqrt{A^2+B^2}=1\] in your example

Answer 5

\[\sin \left( x+\frac{ 7 \pi }{ 4 } \right)=\sin x \cos \frac{ 7 \pi }{ 4 }+\cos x \sin \frac{ 7 \pi }{ 4 }\]
\[=\sin x \cos \left( 2 \pi-\frac{ \pi }{ 4 } \right)+\cos x \sin \left( 2 \pi-\frac{ \pi }{ 4 } \right)\]
\[=\sin x \cos \left( -\frac{ \pi }{ 4 } \right)+\cos x \sin \left( -\frac{ \pi }{ 4 } \right)\]
\[=\sin x \cos \frac{ \pi }{ 4 }-\cos x \sin \frac{ \pi }{ 4 }\]
\[=\frac{ 1 }{ \sqrt{2} }\sin x-\frac{ 1 }{ \sqrt{2} }\cos x \]