N2023 P1 Q6
N2023 P1 Q6
Junior College 2
8 marks
Free
- Using double angle formulae, show that \(\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3)\).[2]
- The region \(R\) lies in the first quadrant and is bounded by the curve \(y^4=(9-x^2)^3\), the \(x\)-axis and the lines \(x=1.5\) and \(x=3\). \(R\) is rotated about the \(x\)-axis through \(2\pi\) radians. Using the substitution \(x=3\sin\theta\), find the exact volume generated.[6]
Default solution
- Write \(c=\cos\theta\), so \(\cos2\theta=2c^2-1\).\(\cos4\theta\)\({}=2\cos^2 2\theta-1\)\({}=2(2c^2-1)^2-1\)\({}=8c^4-8c^2+1\)\(8c^4\)\({}=\cos4\theta+8c^2-1\)\({}=\cos4\theta+4\cos2\theta+3\)\(\therefore\ \cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3)\) (shown).
- In the first quadrant, \(y^2=(9-x^2)^{\frac32}\). The region and axis of rotation are:\(V\)\({}=\pi\int_{\frac32}^{3}(9-x^2)^{\frac32}\,\mathrm{d}x\)Put \(x=3\sin\theta\), with \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{2}\) and \(\mathrm{d}x=3\cos\theta\,\mathrm{d}\theta\). On this interval, \(\sqrt{9-x^2}=3\cos\theta\).\(V\)\({}=81\pi\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cos^4\theta\,\mathrm{d}\theta\)By (a), use \(J_r=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cos(r\theta)\,\mathrm{d}\theta\) for \(r=2,4\).\(J_4\)\({}=\left[\frac{\sin4\theta}{4}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\)\({}=-\frac{\sqrt3}{8}\)\(J_2\)\({}=\left[\frac{\sin2\theta}{2}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\)\({}=-\frac{\sqrt3}{4}\)\(3\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\pi\).\(V\)\({}=\frac{81\pi}{8}(J_4+4J_2+\pi)\)\({}=\frac{81\pi}{8}\left(\pi-\frac{9\sqrt3}{8}\right)\)\({}=\frac{81\pi}{64}(8\pi-9\sqrt3)\)The exact volume is \(\frac{81\pi}{64}(8\pi-9\sqrt3)\) cubic units.
Answer:(a) \(\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3)\) (shown). (b) \(\frac{81\pi}{64}(8\pi-9\sqrt3)\) cubic units.
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