given: a. sin(x) b. cos(x) c. cos(y) d. sin(y) use the drop - down menus in the derivation of the sine sum…

given: a. sin(x) b. cos(x) c. cos(y) d. sin(y) use the drop - down menus in the derivation of the sine sum identity: cos(x + y)=sin((π/2)-(x + y))=sin(((π/2)-x)-y)=sin((π/2)-x)cos(y)-cos((π/2)-x) = cos(y)-sin(x)sin(y)

given: a. sin(x) b. cos(x) c. cos(y) d. sin(y) use the drop - down menus in the derivation of the sine sum identity: cos(x + y)=sin((π/2)-(x + y))=sin(((π/2)-x)-y)=sin((π/2)-x)cos(y)-cos((π/2)-x) = cos(y)-sin(x)sin(y)

Answer

Explanation:

Step1: Recall the sine - difference identity

The sine - difference identity is $\sin(A - B)=\sin(A)\cos(B)-\cos(A)\sin(B)$. Here $A=\frac{\pi}{2}-x$ and $B = y$. So, $\sin((\frac{\pi}{2}-x)-y)=\sin(\frac{\pi}{2}-x)\cos(y)-\cos(\frac{\pi}{2}-x)\sin(y)$.

Step2: Use the co - function identities

We know that $\sin(\frac{\pi}{2}-\alpha)=\cos(\alpha)$ and $\cos(\frac{\pi}{2}-\alpha)=\sin(\alpha)$. So, $\sin(\frac{\pi}{2}-x)=\cos(x)$ and $\cos(\frac{\pi}{2}-x)=\sin(x)$. Then $\sin(\frac{\pi}{2}-x)\cos(y)-\cos(\frac{\pi}{2}-x)\sin(y)=\cos(x)\cos(y)-\sin(x)\sin(y)$.

Answer:

The first blank should be filled with $\sin(y)$ and the second blank should be filled with $\cos(x)$. So the answers are D for the first blank and B for the second blank.