Show that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3 \theta - 3\cos \theta. Hence, or otherwise, evaluate ∫0π/6sin⁡2θcos⁡3θ dθ\displaystyle \int_0^{\pi/6} \sin 2\theta \cos 3\theta\, d\theta exactly.

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Worked solution

2(a)

1. Split $3\theta$ into $(2\theta + \theta)$ and apply the cosine compound angle formula.

cos⁡3θ=cos⁡(2θ+θ)\cos 3\theta = \cos(2\theta + \theta)
=cos⁡2θcos⁡θ−sin⁡2θsin⁡θ= \cos 2\theta \cos \theta - \sin 2\theta \sin \theta

2. Apply the double angle formulas to express the equation in terms of single angles.

=(2cos⁡2θ−1)cos⁡θ−(2sin⁡θcos⁡θ)sin⁡θ= (2\cos^2 \theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta
=2cos⁡3θ−cos⁡θ−2sin⁡2θcos⁡θ= 2\cos^3 \theta - \cos \theta - 2\sin^2 \theta \cos \theta

3. Use the Pythagorean identity to convert all sine terms to cosine and simplify to the final form.

=2cos⁡3θ−cos⁡θ−2(1−cos⁡2θ)cos⁡θ= 2\cos^3 \theta - \cos \theta - 2(1 - \cos^2 \theta)\cos \theta
=2cos⁡3θ−cos⁡θ−2cos⁡θ+2cos⁡3θ= 2\cos^3 \theta - \cos \theta - 2\cos \theta + 2\cos^3 \theta
=4cos⁡3θ−3cos⁡θ= 4\cos^3 \theta - 3\cos \theta

Answer

\text{Shown}

2(b)

1. Substitute the identity from part (a) and the double angle formula for $\sin 2\theta$ into the integral, then expand.

∫0π/6sin⁡2θcos⁡3θ dθ=∫0π/6(2sin⁡θcos⁡θ)(4cos⁡3θ−3cos⁡θ) dθ\int_0^{\pi/6} \sin 2\theta \cos 3\theta \, d\theta = \int_0^{\pi/6} (2\sin \theta \cos \theta)(4\cos^3 \theta - 3\cos \theta) \, d\theta
=∫0π/6(8sin⁡θcos⁡4θ−6sin⁡θcos⁡2θ) dθ= \int_0^{\pi/6} (8\sin \theta \cos^4 \theta - 6\sin \theta \cos^2 \theta) \, d\theta

2. Integrate the expression using the reverse chain rule for powers.

Using ∫[f(x)]nf′(x) dx=[f(x)]n+1n+1+c\int [\mathrm{f}(x)]^n \mathrm{f}'(x) \, dx = \frac{[\mathrm{f}(x)]^{n+1}}{n+1} + c:
=[−85cos⁡5θ+2cos⁡3θ]0π/6= \left[ -\frac{8}{5}\cos^5 \theta + 2\cos^3 \theta \right]_0^{\pi/6}

3. Substitute the upper limit $\frac{\pi}{6}$ into the antiderivative.

At θ=π6:−85(32)5+2(32)3\text{At } \theta = \frac{\pi}{6}: \quad -\frac{8}{5}\left(\frac{\sqrt{3}}{2}\right)^5 + 2\left(\frac{\sqrt{3}}{2}\right)^3
=−85(9332)+2(338)= -\frac{8}{5}\left(\frac{9\sqrt{3}}{32}\right) + 2\left(\frac{3\sqrt{3}}{8}\right)
=−9320+15320=6320=3310= -\frac{9\sqrt{3}}{20} + \frac{15\sqrt{3}}{20} = \frac{6\sqrt{3}}{20} = \frac{3\sqrt{3}}{10}

4. Substitute the lower limit $0$ and subtract to find the exact value.

At θ=0:−85(1)5+2(1)3=−85+2=25\text{At } \theta = 0: \quad -\frac{8}{5}(1)^5 + 2(1)^3 = -\frac{8}{5} + 2 = \frac{2}{5}
Value=3310−25=33−410\text{Value} = \frac{3\sqrt{3}}{10} - \frac{2}{5} = \frac{3\sqrt{3} - 4}{10}

Answer

33−410\dfrac{3\sqrt{3} - 4}{10}

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