Question

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


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Answer 1

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

Explanation:

I will work with radians.

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

First, I will deal with the numerator

`$\cos^2 \left((\pi)/(2)-x \right)+\sin(-x)-\sin^2 \left((\pi)/(2)-x \right)+\cos \left((\pi)/(2)-x \right)$`

Consider the following trigonometric identities:

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

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

`\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} $`