# Proof incorrect

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## Main Question or Discussion Point

What is wrong with my proof?

Let $$\theta=cos^{-1}x-\frac{\pi}{2}$$

Then $$cos\theta=cos\left(cos^{-1}x-\frac{\pi}{2}\right)$$

$$RHS=xcos\frac{\pi}{2}-sin(cos^{-1}x)sin\frac{\pi}{2}$$

$$=-\sqrt{1-x^2}$$

Therefore $$\theta=cos^{-1}(-\sqrt{1-x^2})$$

$$\theta=\pi-cos^{-1}\sqrt{1-x^2}$$

Hence $$cos^{-1}x-\frac{\pi}{2}=\pi-cos^{-1}\sqrt{1-x^2}$$

So finally, $$cos^{-1}x+cos^{-1}\sqrt{1-x^2}=\frac{3\pi}{2}$$

Except this is untrue for all values except $x=-1$. I'm guessing I probably made a substitution which is valid for only certain values. Inverse trig seems to do that a lot to me

Last edited:

Let $$\theta=cos^{-1}x-\frac{\pi}{2}$$

Then $$cos\theta=cos\left(cos^{-1}x-\frac{\pi}{2}\right)$$

$$RHS=xcos\frac{\pi}{2}-sin(cos^{-1}x)sin\frac{\pi}{2}$$
The last line here is wrong. The appropriate identity states:
$$\cos(\alpha-\beta) = \cos\alpha\,\cos\beta + \sin \alpha\,\sin\beta$$
So you should have gotten:
$$RHS=xcos\frac{\pi}{2}+sin(cos^{-1}x)sin\frac{\pi}{2}$$

In the future a good way to identify an error in an argument about trigonometric identities is to simply plug in a number. Preferably one whose value you can calculate under the functions you are working with and which doesn't have very nice symmetric properties since that is usually where errors creep in. For instance in your case I would have tested with x=1/2 which would have yielded:
$$\theta = \cos^{-1}1/2 - \pi/2 = -\pi/6$$
$$\cos\theta = \sqrt{3}\pi/2$$
$$(1/2)\cos\pi/2 - \sin(\cos^{-1}(1/2))\sin\pi/2 = - \sin(\pi/3) = -\sqrt{3}\pi/2$$