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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.