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What is the value of $\cot70+4\cos70?$

So far I expressed $\cot70$ as $\frac{\cos70}{\sin70}$ and $4\cos70$ as $\frac{4\cos70\cdot\sin70}{\sin70}$. I then used the sine double angle formula to get: $\frac{4\cos70\cdot\sin70}{\sin70}=\frac{2\sin140}{\sin70}$ but I'm struggling to find a way to finish this.

Silverleaf1
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    I like Atticus's answer a lot better than the accepted answer on the other question. – saulspatz Feb 23 '20 at 19:48
  • @saulspatz: The identical question has been asked and answered before, therefore it should be closed as a duplicate (so that all solutions are found at one place). One can always add a better answer to the original question (however, in this case there is already a quite similar solution: https://math.stackexchange.com/a/1202958/42969). – Martin R Feb 23 '20 at 19:56
  • @MartinR I was just making a comment; I wasn't making a suggestion or taking issue with you. – saulspatz Feb 23 '20 at 20:02

1 Answers1

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Just using sum to product formulas:

$$\begin{aligned} \cot70^\circ+4\cos 70^\circ &=\dfrac{\cos70^\circ+4\cos70^\circ\sin70^\circ}{\sin70^\circ}\\ &= \dfrac{\sin20^\circ+2\sin140^\circ}{\cos20^\circ}\\ &= \dfrac{\sin20^\circ+2\sin 40^\circ}{\cos20^\circ}\\ &= \dfrac{\sin20^\circ+\sin40^\circ+\cos50^\circ}{\cos20^\circ}\\ &= \dfrac{2\sin30^\circ\cos10^\circ+\cos50^\circ}{\cos20^\circ}\\ &= \dfrac{\cos10^\circ+\cos50^\circ}{\cos20^\circ}\\ &= \dfrac{2\cos20^\circ\cos30^\circ}{\cos20^\circ}\\ &=\sqrt{3}\\ \end{aligned} $$

LHF
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