Answer:
[tex]$\frac{p^2 - 16} {4p^2 + 16} $[/tex]
Step-by-step explanation:
I will work with radians.
[tex]$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$[/tex]
First, I will deal with the numerator
[tex]$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$[/tex]
Consider the following trigonometric identities:
[tex]$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$[/tex]
[tex]$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$[/tex]
[tex]\boxed{\sin(-x)=-\sin(x)}[/tex]
[tex]\boxed{\cos(-x)=\cos(x)}[/tex]
Therefore, the numerator will be
[tex]$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$[/tex]
Once
[tex]\sin(x)=p[/tex]
[tex]\cos(x)=4[/tex]
[tex]$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$[/tex]
Now let's deal with the numerator
[tex][\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)][/tex]
Using the sum and difference identities:
[tex]\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}[/tex]
[tex]\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}[/tex]
[tex]\sin(\pi -x) = \sin(x)[/tex]
[tex]\sin(2\pi +x)=\sin(x)[/tex]
[tex]\cos(2\pi-x)=\cos(x)[/tex]
Therefore,
[tex][\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)][/tex]
[tex]\implies [p+4] \cdot [p \cdot 4]=4p^2+16p[/tex]
The final expression will be
[tex]$\frac{p^2 - 16} {4p^2 + 16} $[/tex]
