I'm learning about trig identities, and I'm struggling to prove that two expressions are equal:

$$ \frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B} $$

How do you go about proving this? I know about compound angles - i.e. the sine, cosine and tangent of $(A \pm B)$, but don't know how to apply it in this situation.

  • 8
    $\begingroup$ the ' at the end of the denominator in your equation is incorrect. This is simply a comma , in the original printed formula. $\endgroup$ – Pixel Feb 8 '14 at 17:28

Using the tangent addition formulas, we get

$$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$


$$\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

From this, we get

$$\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{1 - \tan A \tan B}{1 + \tan A \tan B} = \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B + \sin A \sin B}$$

$$= \frac{\tan (A+B)}{\tan (A-B)} \frac{\cos (A+B)}{\cos (A-B)} = \frac{\sin (A+B)}{\sin (A-B)}.$$

  • 2
    $\begingroup$ How does $\frac{1 - tanAtanB}{1 + tanAtanB} = \frac{cosAcosB - sinAsinB}{cosAcosB + sinAsinB}$? $\endgroup$ – hohner Feb 8 '14 at 17:34
  • 3
    $\begingroup$ @hohner multiply top and bottom by cosAcosB $\endgroup$ – oks Feb 8 '14 at 17:38
  • 1
    $\begingroup$ I see it now. Thanks guys $\endgroup$ – hohner Feb 8 '14 at 17:39

We start with $\frac{\tan A + \tan B}{\tan A - \tan B}$:

$$\frac{\tan A + \tan B}{\tan A - \tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}}=\frac{\sin A\cos B +\sin B\cos A}{\sin A\cos B -\sin B\cos A}=\frac{\sin(A+B)}{\sin(A-B)}.$$


$$ \frac{\sin(A + B)}{\sin(A - B)}=\frac{\sin A\cos B+\sin B\cos A}{\sin A\cos B-\sin B\cos A}=\frac{\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}}{\frac{\sin A\cos B-\sin B\cos A}{\cos A\cos B}}=\frac{\tan A + \tan B}{\tan A - \tan B} $$


Using the fact that $e^{i x}=\cos x + i\sin x$, where $i=\sqrt{-1}$, we have

$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{\Im(e^{iA}e^{iB})}{\Im(e^{iA}e^{-iB})} = \frac{\Im((\cos A+i\sin A)(\cos B+i\sin B))}{\Im((\cos A+i\sin A)(\cos B-i\sin B))},$$

where $\Im$ denotes the imaginary part. Expanding and taking the imaginary ($\Im$) parts, we get

$$\frac{\cos A\sin B+\sin A\cos B}{\sin A\cos B-\cos A\sin B}.$$

Dividing numerator and denominator by $\cos A \cos B$ gives

$$\frac{\left(\displaystyle\frac{\cos A\sin B+\sin A\cos B}{\cos A\cos B}\right)}{\left(\displaystyle\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}\right)} = \frac{\tan B+\tan A}{\tan A-\tan B},$$

as required.


A very general (but extremely useful) approach is by noting the following

$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$ $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

and since $\tan(x) = \frac{\sin(x)}{\cos(x)}$

$$\tan(x) = \frac{1}{i}\frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} = -i \frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} $$

We note here that $i$ is the imaginary constant (basically its a number such that $i^2 = -1$) I will leave you to go ahead and verify these formulas work for every trig Identity you have already memorized and more information is underneath: http://en.wikipedia.org/wiki/Euler%27s_formula

So we wish to prove:

$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{\tan(A) + \tan(B)}{\tan(A) - \tan(B)} $$

We prepare the left side (noting that both fractions take form $\frac{A}{2i}$ and therefore we can drop the $2i$ denominators

$$\frac{\sin(A+B)}{\sin(A-B)} = \frac{e^{i(A+B)} - e^{-i(A+B)}}{e^{i(A-B)} - e^{-i(A-B)}}$$

So because I'm slightly lazy (and for the sake of giving you something to practice) you need to do the exact same job with the tan(x) expression where each instance of x becomes A or B depending on whats being evaluated.

Now the goal is to systematically simplify both expressions by transforming expressions of the form $e^{-k}$ to $\frac{1}{e^k}$

Followed by taking sums and giving them common denominators $\frac{A}{C} + \frac{B}{D} = \frac{AD + BC}{CD}$

And dividing out common factors.

Its a tedious process but once done. Both expression will look exactly the same... Well you don't have to take my word for it, do it yourself and prove that it works ;)


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.