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# Prove that $\cos (\sin^{-1} x) = \sqrt{1 - x^2}$.

## $$\text { Let } y=\sin ^{-1} x . \text { Then }-\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \Rightarrow \text { cos } y \geq 0, \operatorname{so~} \cos \left(\sin ^{-1} x\right)=\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-x^{2}}$$.

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### Video Transcript

here we have a statement that we want to prove and I like to do a graphically with reference triangles. I think we need to consider three cases. Case one X is greater than zero. So x is positive. Case two X is less than zero negative. In case three the only remaining possibility X equals zero. So imagine that X is positive. So the inverse sign of a positive means We must be looking at a reference strangling Quadra one. And if the sine of the angle we can call this angle Fada supposed, we can If the sine of the angle is X, then that means the opposite over high pot news would be X over one. From here we use the Pythagorean theorem to find the length of the adjacent side and we will get the square root of one minus X squared. Okay. Now, to finish off the case, we want to find the co sign of angle. Fada and co sign is adjacent over high pot news. So that would be square root one minus x squared over one. And that is the square root of one minus X squared. Okay, Now suppose that X is negative, less than zero. So that means we would have a reference triangle in quadrant four. So our angle fada is this one outside the triangle. And still the sign is X. So the opposite is X and the high pod news is one. When we use the Pythagorean theorem would get the same thing we got in the last case for the co sign for the adjacent side, we get one minus X squared. Now the coastline of the angle, just like it was last time, is going to be adjacent over high pot news. So again we get this queried of one minus X squared. Now suppose x zero if x zero, then that means inverse sign of X must be zero. The sign of zero radiance is zero. So do that backwards. You still zero. And now what's The coastline of that coastline of zero is one, and that is the same as the square root of one minus zero squared. That's also one

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