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I need to prove:

$$ \frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x} = \tan2x $$

The sum and product formulae are relevant:

$$ \sin(A + B) + \sin (A-B) = 2 \sin A \cos B \\ \sin(A + B) - \sin (A-B) = 2 \cos A \sin B \\ \cos(A + B) + \cos (A-B) = 2 \cos A \cos B $$

Taking the numerator first:

$$ \sin x + \sin 2x + \sin3x = 2\sin\left(\frac{3}{2}x\right) \cos\left(\frac{1}{2}x\right) + \sin 3x \\ \sin 3x = \sin\left(\frac{3}{2}x + \frac{3}{2}x\right) - \sin\left(\frac{3}{2}x - \frac{3}{2}x\right) = 2\cos\left(\frac{3}{2}x\right)\sin\left(\frac{3}{2}x\right) \\ \therefore \sin x + \sin 2x + \sin3x=2\sin\left(\frac{3}{2}x\right)\left[\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{3}{2}x\right)\right] $$

and now the the denominator:

$$ \cos x + \cos 2x + \cos 3x = 2\cos\left(\frac{3}{2}x\right) \cos\left(\frac{1}{2}x\right) + \cos 3x $$

But I don't know how to express $\cos3x$ in terms of a product, so I can't factorize and cancel. Can someone help me make the next few steps?

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1 Answer 1

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\begin{align} \frac{\sin x + \sin 2x + \sin3x}{\cos x + \cos 2x + \cos 3x}&=\frac{\sin x + \sin 3x + \sin2x}{\cos x + \cos 3x + \cos 2x}\\ &=\frac{2\sin 2x\cos x+ \sin 2x}{2\cos 2x\cos x + \cos 2x}\\ &=\frac{\sin2x(2\cos x+1)}{\cos 2x(2\cos x + 1)}\\ & = \tan2x \end{align}

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