Question

using a double - angle or half - angle formula to simplify the given expressions.(a) if $\cos^{2}(30^{\circ})-\sin^{2}(30^{\circ}) = \cos(a^{\circ})$, then $a =$ degrees.(b) if $\cos^{2}(3x)-\sin^{2}(3x)=\cos(b)$, then $b =$.question help: worked example 1

Explanation:

Step1: Recall the double - angle formula for cosine

The double - angle formula for cosine is \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta\).

Step2: Solve for \(A\) in part (a)

Given \(\cos^{2}(30^{\circ})-\sin^{2}(30^{\circ})=\cos(A^{\circ})\).
By the double - angle formula \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta\), when \(\theta = 30^{\circ}\), we have \(2\theta=A\).
Substitute \(\theta = 30^{\circ}\) into \(2\theta\), then \(A = 2\times30^{\circ}=60^{\circ}\).

Step3: Solve for \(B\) in part (b)

Given \(\cos^{2}(3x)-\sin^{2}(3x)=\cos(B)\).
By the double - angle formula \(\cos(2\theta)=\cos^{2}\theta-\sin^{2}\theta\), when \(\theta = 3x\), we have \(B = 2\times3x=6x\).

Answer:

a. \(60\) degrees
b. \(6x\)