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I require help in the area of trigonometry in proving an identity.

I am to prove that the left hand side is equal to $\tan2 \theta$.

I understand up until the second step in this calculation (original image here):

$$\begin{align*}\text{LHS}&=\frac{\sin 3\theta+\sin\theta}{\cos 3\theta+\cos\theta}\\\\ &=\frac{2\sin\left(\frac{3\theta+\theta}2\right)\cos\left(\frac{3\theta-\theta}2\right)}{2\cos\left(\frac{3\theta+\theta}2\right)\cos\left(\frac{3\theta-\theta}2\right)}\\\\ &=\frac{\sin 2\theta}{\cos2\theta}\\\\ &=\tan2\theta\\\\ &=\text{RHS} \end{align*}$$

How did they jump from the factor formulae to the double angle one?

I cannot see the relation between those two, am I missing something?

Thank you for your help.

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  • $\begingroup$ Which part do you find confusing - $\frac {(3 \theta + \theta)} 2 = 2 \theta$, or the factoring out of the common factors? $\endgroup$ – mathguy Mar 29 '16 at 16:12
  • $\begingroup$ I understand the solution completely now thank you, what do you mean by factoring out common factors though? $\endgroup$ – Ian Mar 29 '16 at 16:23
  • $\begingroup$ I am sure @mathguy means the canceling out of the common factors in the second line (canceling out in the way $\frac{b\cdot a} a =b$) $\endgroup$ – user190080 Mar 29 '16 at 16:26
  • $\begingroup$ Yes, that is what I mean. You have the 2 in the numerator and the denominator, so it cancels out, and the same with the third factor in the numerator and denominator. $\endgroup$ – mathguy Mar 29 '16 at 16:27
  • $\begingroup$ It might seem silly to ask but how come the 2 thetas with the sin and the cos wasn't cancelled out also? $\endgroup$ – Ian Mar 29 '16 at 16:33
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Whats going on is that there is a miniature u-substitution step in between steps 1 and 2 of the proof. If you let x = θ and y = 3θ then use the fact that

$$\sin (x) + \sin (y) = 2 \sin \frac{x + y}{2} \cos \frac{x-y}{2}$$,

then re-substitute the original values then you get the second step of the proof.

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  • $\begingroup$ you could also add a "\" just before the $sin$ to make it look like a proper $\sin$ $\endgroup$ – user190080 Mar 29 '16 at 16:39

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