Q.3 (c) If \(\sin^4\theta\cos^3\theta=a_1\cos\theta+a_3\cos3\theta+a_5\cos5\theta+a_7\cos7\theta\), then prove that \(a_1+9a_3+25a_5+49a_7=0\).
Best answer
Let \(x=\cos\theta+i\sin\theta\): \(x-\frac1x=2i\sin\theta,\ x+\frac1x=2\cos\theta,\ x^n+\frac1{x^n}=2\cos n\theta\).
\((2i\sin\theta)^4(2\cos\theta)^3=\left(x-\frac1x\right)^4\left(x+\frac1x\right)^3=\left(x-\frac1x\right)\left(x^2-\frac1{x^2}\right)^3\)
\(=\left(x-\frac1x\right)\left(x^6-3x^2+\frac3{x^2}-\frac1{x^6}\right)=\left(x^7+\frac1{x^7}\right)-\left(x^5+\frac1{x^5}\right)-3\left(x^3+\frac1{x^3}\right)+3\left(x+\frac1x\right)\)
\(128\sin^4\theta\cos^3\theta=2\cos7\theta-2\cos5\theta-6\cos3\theta+6\cos\theta\)
\(\sin^4\theta\cos^3\theta=\frac1{64}\left[3\cos\theta-3\cos3\theta-\cos5\theta+\cos7\theta\right]\)
So \(a_1=\frac3{64},\ a_3=-\frac3{64},\ a_5=-\frac1{64},\ a_7=\frac1{64}\).
\(a_1+9a_3+25a_5+49a_7=\frac{3-27-25+49}{64}=\frac0{64}\) → \(a_1+9a_3+25a_5+49a_7=0\) (proved).
(Check: differentiating the identity twice and putting \(\theta=0\) gives \(-(a_1+9a_3+25a_5+49a_7)\) = second derivative of \(\sin^4\theta\cos^3\theta\) at 0, which is 0.)
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