I was going through the derivatives of trigonometric and inverse trigonometric functions in my book and I'm just done with the proof of the derivative of arcsin(x) $$ \dfrac{d}{dx} \sin^{-1}x = \dfrac{1}{\cos(y)}$$ were we define $\cos(y)$ in terms of $x$ as $\sqrt{1-x^2}$ so we have the following derivative $$ \dfrac{d}{dx} \sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$$ but I don't really get why it is the case that $\cos^{-1} (x)=\frac{\pi}{2}-\sin^{-1} (x)$ so the derivative for $\cos^{-1} (x)$ is gonna be the same but in a negative sign, is there another way to prove this derivative of arccosine? or can someone explain to me in a simple language why $\cos^{-1} (x)=\frac{\pi}{2}-\sin^{-1} (x)$
Note: I saw an answer but it didn't explain the relation, and other sources online don't really make it look simple
Best Answer
The relationship you mentioned between ($\cos^{-1}(x)$) & ($\sin^{-1}(x)$) involves the properties of complementary angles & it's not hard tbh it's really simple when you think about it, just re-draw the pic down here while reading so you can have a better imagination about what I'm saying, name the angle $y$ as $\sin^{-1}(x)$, and then name the second acute angel $\cos^{-1}(x)$
Consider a right-angled triangle with one acute angle, say $y$ since we wanna get $\cos y$ in terms of $x$. If you define $y$ as the angle such that $\sin(y) = x$, then $y= \sin^{-1}(x)$. Now, in this triangle, $\cos(y)$ would be the ratio of the adjacent side to the hypotenuse.
Now, recall that in a right-angled triangle, the sum of the 2 angles -excluding the right angle- is always $\frac{\pi}{2}$ radians (or 90 degrees). Therefore, if $y$ is one of the angles, the other acute angle in this triangle would be $\frac{\pi}{2} - y$. So, if $y= \sin^{-1}(x)$, then the other acute angle, which corresponds to $\cos^{-1}(x)$, would be $\frac{\pi}{2} - y$.
In other terms, $\cos(\frac{\pi}{2} - y) = \sin(y)$, which translates to $\cos(\cos^{-1}(x)) = \sin(\sin^{-1}(x))$. Simplifying this gives $\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$.
Therefore, because of the relationship between the angles in a right-angled triangle, the inverse cosine function of $x$ is equal to $\frac{\pi}{2}$ minus the inverse sine function of $x$. This relationship helps to establish the derivative of $\cos^{-1}(x)$ using the derivative of $\sin^{-1}(x)$ with the negative sign. that's all, it's easy