If \(a \sin^{3}\theta + b \cos^{3}\theta = \sin\theta \cos\theta\) and \(a \sin\theta = b \cos\theta\), then value of \(a^{2} + b^{2}\) is
1
Write \(a\sin^{3}\theta = (a\sin\theta)\sin^{2}\theta\) and \(b\cos^{3}\theta = (b\cos\theta)\cos^{2}\theta\).
Using \(b\cos\theta = a\sin\theta\): \(a\sin^{3}\theta + b\cos^{3}\theta = a\sin\theta(\sin^{2}\theta + \cos^{2}\theta) = a\sin\theta\).
So \(a\sin\theta = \sin\theta\cos\theta\), giving \(a = \cos\theta\) (for \(\sin\theta \neq 0\)).
Then \(b\cos\theta = a\sin\theta = \sin\theta\cos\theta\), giving \(b = \sin\theta\).
Therefore \(a^{2} + b^{2} = \cos^{2}\theta + \sin^{2}\theta = 1\).
Check: at \(\theta = 45^{\circ}\), \(a = b = \frac{1}{\sqrt{2}}\); then \(a\sin^{3}\theta + b\cos^{3}\theta = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{2\sqrt{2}} = \frac{1}{2} = \sin\theta\cos\theta\), and \(a^{2} + b^{2} = \frac{1}{2} + \frac{1}{2} = 1\).
Hence, 1.
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