Note that sin(cos^{-1} x) or sin(arc cosx) are algebraic functions, not trigonometric functions. The value of sin(cos^{-1} x) or sin(arc cosx) is equal to $\sqrt{1-x^2}$. In this post, we will simplify sin(cos^{-1} x).

The formula of sin(cos inverse x) is given below:

sin(cos^{-1} x) = root(1-x^{2}).

## Simplify sin(cos^{-1}(x))

We all know that $\sin(\cos^{-1}(x))= \sin(\text{arc} \cos x)$. To simplify this function, we will follow the below steps.

**Step 1:** In the first step, we let that $\alpha=\cos^{-1} x$ $\cdots (I)$

As $\alpha=\cos^{-1} x$ and we have to simplify $\sin(\cos^{-1}x)$, we need to find out the value of $\sin \alpha$. From the equation $(I)$, we obtain that

$\cos \alpha=x$ $\cdots (II)$

**Step 2:** Now, we use the Pythagorean trigonometric identity $\sin^2 \alpha+\cos^2 \alpha =1$. From this identity, we have that

$\sin^2\alpha = 1-\cos^2 \alpha$

$\Rightarrow \sin \alpha = \sqrt{1-\cos^2 \alpha}$

$=\sqrt{1-x^2}$ [ putting the value of $\cos \alpha$ from $(II)$ ]

$\therefore \sin \alpha =\sqrt{1-x^2}$

**Step 3:** Next, we will put the value of $\alpha$ from $(I)$. By doing so we obtain that

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

**Conclusion:** Thus, the formula of $\sin(\cos^{-1}x)$ is equal to $\sqrt{1-x^2}$.

**Also Read:**

**sinx=0, cosx=0, tanx=0 General Solution**

**Values of sin 15, cos 15, tan 15**

**Values of sin 75, cos 75, tan 75**

**Question 1:** Find the value of sine of cosine inverse 0, that is, find $\sin(\cos^{-1}0)$.

*Answer:*

Using the above formula, we have that $\sin(\cos^{-1}0)=\sqrt{1-0^2}=1$.

**Question 2:** Find the value of sine of cosine inverse 1, that is, find $\sin(\cos^{-1}1)$.

*Answer:*

Using the above formula, we have that $\sin(\cos^{-1} 0)=\sqrt{1-1^2}=0$.

## FAQs

**Q1: What is sin(cos^-1(x))?**

**Answer:** The value of sin(cos^{-1}x) is equal to $\sqrt{1-x^2}$.