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Chapter 1: Problem 63

Prove the following identities. $$\cos ^{-1} x+\cos ^{-1}(-x)=\pi$$

Short Answer

Expert verified
Question: Prove that the sum of the inverse cosine function of x and the inverse cosine function of -x is equal to pi. Answer: To prove that $$\cos ^{-1} x+\cos ^{-1}(-x)=\pi$$, we first expressed the given identity in terms of unknown angles $$\alpha$$ and $$\beta$$. Then we used the inverse cosine definition, cosine addition formula, and Pythagorean identity to find the sine and cosine values for $$\alpha$$ and $$\beta$$. After determining the correct sign for sine values, we applied the cosine addition formula and used the inverse cosine function again to show that the sum of the angles is equal to $$\pi$$.

Step by step solution

01

Recall the definition of inverse trigonometric functions

Inverse trigonometric functions are used to find the angle whose trigonometric function value is given. Here, we are working with the inverse cosine function, which is also called arccos or cos^-1. For any x between -1 and 1, if cos(theta) = x, then cos^-1(x) = theta.
02

Express the given identity with respect to unknown angles

Let's denote the angles associated with the given identity as follows: $$\cos ^{-1} x = \alpha$$ $$\cos ^{-1}(-x) = \beta$$ We want to prove that the $$\alpha + \beta = \pi$$.
03

Utilize the inverse cosine definition

Since we defined the angles $$\alpha$$ and $$\beta$$ as the inverse cosine of x and -x, let's find the cosine values: $$\cos(\alpha) = x$$ $$\cos(\beta) = -x$$
04

Recall the cosine addition formula

The cosine addition formula states that: $$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$ Now we need to find the sine values of $$\alpha$$ and $$\beta$$.
05

Use the Pythagorean identity to find sine values.

The Pythagorean identity states that: $$\sin ^2 \theta + \cos ^2 \theta = 1$$ Substitute the values of $$\alpha$$ and $$\beta$$: $$\sin ^2 \alpha + \cos ^2 \alpha = 1 \Rightarrow \sin(\alpha) = \pm \sqrt{1 - x^2}$$ $$\sin ^2 \beta + \cos ^2 \beta = 1 \Rightarrow \sin(\beta) =\pm \sqrt{1 - (-x)^2} = \pm \sqrt{1 - x^2}$$
06

Determine the correct sign for sine values

Since $$\alpha$$ and $$\beta$$ are in the range of $$[0, \pi]$$, both $$\sin(\alpha)$$ and $$\sin(\beta)$$ are positive. Therefore, $$\sin(\alpha) = \sqrt{1-x^2}$$ $$\sin(\beta) = \sqrt{1-x^2}$$
07

Apply the cosine addition formula

Now, let's find the cosine of $$\alpha + \beta$$ by using the cosine addition formula: $$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$$ Using the values we found for the cosine and sine functions, we have: $$\cos(\alpha + \beta) = x(-x) - \sqrt{1 - x^2}\sqrt{1 - x^2} = -x^2 - (1 - x^2) = -1$$
08

Use the inverse cosine function again

We know that the cosine of $$\pi$$ is equal to -1: $$\cos(\pi) = -1$$ Now, we see that $$\cos(\alpha + \beta) = -1$$. This means that: $$\alpha + \beta = \cos^{-1}(-1) = \pi$$ And we have successfully proven that $$\cos ^{-1} x+\cos ^{-1}(-x)=\pi$$.

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