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Sums Over Sums

Part of the Trig Identities section of Coddy's Trigonometry journey. Lesson 41 of 48.

A fraction with a sum on top and a sum below often turns into two products that share a factor. Take sin⁡(5x)+sin⁡(3x)cos⁡(5x)+cos⁡(3x)\frac{\sin\left(5x\right)+\sin\left(3x\right)}{\cos\left(5x\right)+\cos\left(3x\right)}. The top is 2sin⁡(4x)cos⁡(x)2\sin\left(4x\right)\cos\left(x\right) and the bottom is 2cos⁡(4x)cos⁡(x)2\cos\left(4x\right)\cos\left(x\right). The 2 and the cos x cancel, and sin⁡(4x)cos⁡(4x)\frac{\sin\left(4x\right)}{\cos\left(4x\right)} is tan⁡(4x)\tan\left(4x\right).

Turn both into products first, then look for the common factor. A sum cannot be cancelled term by term; a product can.

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Challenge

Prove that sin⁡(3x)+sin⁡(x)cos⁡(3x)+cos⁡(x)\frac{\sin\left(3x\right)+\sin\left(x\right)}{\cos\left(3x\right)+\cos\left(x\right)} equals tan⁡(2x)\tan\left(2x\right): rewrite it until the board reads tan⁡(2x)\tan\left(2x\right). Then press Check.

Try it yourself

sin⁡3x+sin⁡xcos⁡3x+cos⁡x\frac{\sin 3x+\sin x}{\cos 3x+\cos x}

Rewrite as tan⁡2x\tan 2x

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