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Chapter 3: Problem 10

Prove the given identity. $$ \sin ^{4} \theta-\cos ^{4} \theta=\sin ^{2} \theta-\cos ^{2} \theta $$

Short Answer

Expert verified
\( \sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta \)

Step by step solution

01

Recognize and Use the Difference of Squares

The left hand side of the equation is \( \sin^{4} \theta - \cos^{4} \theta \). Recognize that this can be factored using the difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \). Here, \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Rewrite the equation as \( (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \).
02

Simplify Using a Pythagorean Identity

Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( 1 \) for \( \sin^2 \theta + \cos^2 \theta \) in the factored expression. This gives \( (\sin^2 \theta - \cos^2 \theta)(1) \).
03

Simplify the Expression

Simplify the expression further. Since multiplying by 1 does not change the value, the expression simplifies to \( \sin^2 \theta - \cos^2 \theta \).
04

Verify Both Sides of the Equation

Notice that the simplified expression \( \sin^2 \theta - \cos^2 \theta \) is exactly the right hand side of the original equation. Therefore, the identity is proven: \( \sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta \).

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Most popular questions from this chapter

Suppose that you are given a system of two equations of the following form: $$ \begin{array}{c} A \cos \phi=B \nu_{1}-B \nu_{2} \cos \theta \\ A \sin \phi=B \nu_{2} \sin \theta \end{array} $$ Show that \(A^{2}=B^{2}\left(\nu_{1}^{2}+\nu_{2}^{2}-2 \nu_{1} \nu_{2} \cos \theta\right)\).

Use \(75^{\circ}=45^{\circ}+30^{\circ}\) to find the exact value of \(\sin 75^{\circ}\).

Find the exact values of \(\sin (A+B)\), \(\cos (A+B)\), and tan \((A+B)\). \( \sin A=\frac{8}{17}, \cos A=\frac{15}{17}, \sin B=\frac{24}{25}\), \(\cos B=\frac{7}{25}\)

In general, what is the largest value that \(\sin \theta \cos \theta\) can take? Justify your answer.

Prove the given identity. $$ \frac{\cos ^{2} \psi}{\cos ^{2} \theta}=\frac{1+\cos 2 \psi}{1+\cos 2 \theta} $$
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