Question

If sinx = p and cosx = 4, work out the following forms :​

Answer 1

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$