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Prove that $$\sin\theta+\sin2\theta+\sin3\theta+\cdots+\sin n\theta=\frac{\sin\frac{(n+1)\theta}2\sin\frac{n\theta}2}{\sin\frac\theta2},n\ge1,n\in\mathbb Z.$$

I have solved the base case $n = 1$ and prepared the assumption $n = k$. After I substitute the assumption into the induction step at $n = k+1$ I get lost, as I don't know how to successfully manipulate the numerator into the RHS. Any hints?

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2 Answers 2

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Here is what I have done so far. I cant think of a way to manipulate the LHS into the RHS

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  • $\begingroup$ In the comments of my hint, I have tried to explain how it completes this proof. $\endgroup$
    – PolymorphismPrince
    Commented Nov 23, 2020 at 5:04
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BIG HINT

(Warning: I've written the entire inductive step, just cryptically)

$$\sin{0.5(n+2)}\theta\sin{0.5(n+1)\theta}-\sin{0.5(n+1)}\theta\sin{0.5n\theta}$$

$$=\sin{0.5(n+1)\theta}(\sin{0.5(n+2)}\theta-\sin{0.5n\theta})$$

$$=2\sin{0.5(n+1)\theta}\cos{\frac{0.5(n+2)\theta+0.5n\theta}{2}}\sin{\frac{0.5(n+2)\theta-0.5n\theta}{2}}$$

$$=2\sin{0.5(n+1)\theta}\cos{0.5(n+1)\theta}\sin{0.5\theta}$$

$$=\sin(n+1)\theta \sin{0.5\theta}$$

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  • $\begingroup$ ??? Where do we get the (n+2) if we are assuming n = k+1 (or just n => n+1) $\endgroup$
    – user71207
    Commented Nov 21, 2020 at 7:09
  • $\begingroup$ I posted my attempt as an answer - would you be able to have a look? $\endgroup$
    – user71207
    Commented Nov 21, 2020 at 7:29
  • $\begingroup$ @user71207 to understand what I have done in my hint, equate the first line with last line. Then add $\sin{0.5(n+1)\theta\sin{0.5n\theta}}$ to both sides and divide by $\sin{0.5\theta}$. Hopefully you will see then that I have shown exactly the algebra you need to complete your proof. $\endgroup$
    – PolymorphismPrince
    Commented Nov 23, 2020 at 5:03

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