sin(90° + x) + sin(90° - x) = (sin 90° cos x + cos 90° sin x) + ((sin ) cos x - cos 90° sin x)\n= 2(sin )…

Explanation:

Step1: Recall the angle - addition formula for sine

The formula for $\sin(A + B)=\sin A\cos B+\cos A\sin B$ and $\sin(A - B)=\sin A\cos B-\cos A\sin B$. For $\sin(90^{\circ}+x)=\sin90^{\circ}\cos x+\cos90^{\circ}\sin x$ and $\sin(90^{\circ}-x)=\sin90^{\circ}\cos x-\cos90^{\circ}\sin x$. So the first blank should be $90$.

Step2: Combine the two expressions

$\sin(90^{\circ}+x)+\sin(90^{\circ}-x)=(\sin90^{\circ}\cos x+\cos90^{\circ}\sin x)+(\sin90^{\circ}\cos x - \cos90^{\circ}\sin x)=2\sin90^{\circ}\cos x$. So the second blank is also $90$.

Step3: Evaluate $\sin90^{\circ}$

We know that $\sin90^{\circ}=1$. So $2\sin90^{\circ}\cos x = 2\times1\times\cos x=2\cos x$. The third blank is $1$.

Answer:

First blank: $90$; Second blank: $90$; Third blank: $1$