The inverse cosine function is represented as $f(x)=cos^{-1}x$. Here the -1 is not an exponent, it represents the inverse.The another name for $cos^{-1}x$ is arccosx.

ask-math has already explains you about the sine graph. The domain for sine is $\frac{-\prod }{2}\leq x \leq \frac{\prod}{2}$. The cosine inverse function is not one-to-one function so it is restricted sine function.

The another name for inverse sine function is :

To evaluate inverse cosine functions remember that the following statement is equivalent.

In other words, when we evaluate cosine inverse function we are asking what angle,θ, did we plug into the cosine function (regular, not inverse!) to get x.

f(x)= cos(x) |
Domain ${-\infty}< x < {\infty}$ |
Range $-1\leq y \leq 1$ |

$f(x)=cos^{-1}x$ |
$-1\leq x \leq 1$ |
$0 \leq y \leq {\prod}$ |

In cosine function, we know that $cos \Theta =\frac{adjacent side }{hypotenuse}$

The inverse sine function $cos{-1}$ takes the ratio adjacent/hypotenuse and gives the angle θ.

∴ θ = $cos^{-1}(\frac{adjacent side }{hypotenuse})$

For example, $cos 60^{0} =\frac{1}{2}$

∴ θ = $cos^{-1}(\frac{1}{2})$

⇒ θ = $60^{0}$

1) Evaluate f(x) = $cos^{-1}(0)$

y = $cos^{-1}(0)$

⇒ y = $90^{0}$ ( For the degree you can use a unit circle or calculator)

2) Evaluate f(x) = $cos^{-1}(\frac{\sqrt{3}}{2})$

y = $cos^{-1}(\frac{\sqrt{3}}{2})$

⇒ y = $30^{0}$

3) Find y, when cos(y) = 0.1276.

∴ y = $cos^{-1}(0.1276) $

⇒ y = $82.6690^{0}$

⇒ y = $82.7^{0}$

4) Evaluate : f(x) = $cos^{-1}(cos(\frac{\prod }{4}))$

y = $cos^{-1}(cos(\frac{\prod }{4}))$

The value of $cos(\frac{\prod }{4}) = (\frac{\sqrt{2}}{2})$

∴ y = $cos^{-1}(\frac{\sqrt{2}}{2})$

∴ y = $45^{0}$

5) Evaluate f(x) = $cos^{-1}(-1)$

y = $cos^{-1}(-1)$

⇒ y = $180^{0}$

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