Chapter 9: Problem 19

Show \(\cos 3 A=4 \cos ^{3} A-3 \cos A\)

Short Answer

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Question: Prove the trigonometric identity \(\cos 3A = 4\cos^3 A - 3\cos A\). Answer: By expressing \(\cos 3A\) as \(\cos(2A + A)\) and using double angle formulas, we transformed the expression step by step and found that \(\cos 3A = 4\cos^3 A - 3\cos A\), thus proving the identity.

Step by step solution

Express \(\cos 3A\) as \(\cos(2A + A)\)

Using the formula for the cosine of a sum of angles, we have: \[\cos 3A = \cos(2A + A) = \cos 2A \cos A - \sin 2A \sin A\]

Express \(\cos 2A\) and \(\sin 2A\) using double angle formulas

The double angle formulas are: \[\cos 2A = 2\cos^2 A - 1\] \[\sin 2A = 2\sin A \cos A\] Using these formulas, we can replace \(\cos 2A\) and \(\sin 2A\) from Step 1: \[\cos 3A = (2\cos^2 A - 1)\cos A - (2\sin A \cos A)(\sin A)\]

Substitute \(\sin^2 A = 1 - \cos^2 A\)

Now, we need to use the identity \(\sin^2 A = 1 - \cos^2 A\) to rewrite the term with \(\sin A\) in Step 2: \[\cos 3A = (2\cos^2 A - 1)\cos A - 2(1 - \cos^2 A)(\cos A)(\sin A)\]

Simplify the expression

We will now simplify the expression obtained in Step 3 by distributing and combining like terms: \begin{align*} \cos 3A &= 2\cos^3 A - \cos A - 2\cos A \sin A + 2\cos^3 A \sin A \\ &= 2\cos^3 A - \cos A - 2\cos A (1 - \cos^2 A) \\ &= 2\cos^3 A - \cos A - 2\cos A + 2\cos^3 A \\ &= 4\cos^3 A - 3\cos A \end{align*} Therefore, we have shown that \(\cos 3A = 4\cos^3 A - 3\cos A\), as required.

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Show \(\cos 3 A=4 \cos ^{3} A-3 \cos A\)

Short Answer

Step by step solution

Express \(\cos 3A\) as \(\cos(2A + A)\)

Express \(\cos 2A\) and \(\sin 2A\) using double angle formulas

Substitute \(\sin^2 A = 1 - \cos^2 A\)

Simplify the expression

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