Discussion
If \(\tan \theta = \frac{a}{b}\), then \(b\cos 2\theta + a\sin 2\theta\) is equal to:
Answer: B
We use the double angle formulas in terms of tan θ:
\(\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}\) and \(\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}\)
Substitute \(\tan \theta = a/b\):
\(\cos 2\theta = \frac{1-a^2/b^2}{1+a^2/b^2} = \frac{b^2-a^2}{b^2+a^2}\)
\(\sin 2\theta = \frac{2(a/b)}{1+a^2/b^2} = \frac{2ab}{a^2+b^2}\)
The expression is \(b(\frac{b^2-a^2}{a^2+b^2}) + a(\frac{2ab}{a^2+b^2}) = \frac{b^3-a^2b+2a^2b}{a^2+b^2} = \frac{b^3+a^2b}{a^2+b^2} = \frac{b(b^2+a^2)}{a^2+b^2} = b\).
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