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Soroban · #2 October 3rd, 2007, 12:52 PM

Hello, Tom G!

Quote:

I've got to re-write $3\sin x\cos x + 4\cos^2\!x$
in the form $a\sin2x + b\sin2x +c$ . . . . This can't be right!
. . Do you mean: . ${\color{blue}a\sin^2\!x + b\sin2x + c}$

You're expected to know these two identities: . $\begin{array}{ccc}2\sin\theta\cos\theta & = & \sin2\theta\\ \cos^2\!\theta & = & 1 - \sin^2\!\theta\end{array}$

Then: . $3\sin x\cos x + 4\cos^2\!x \;=\;\frac{3}{2}(2\sin x\cos x) + 4\left(1 - \sin^2\!x\right)$

. . $= \;\frac{3}{2}\sin2x + 4 - 4\sin^2\!x \;=\; -4\sin^2\!x + \frac{3}{2}\sin2x + 4$