# Proving a trig identity

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.

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the ' at the end of the denominator in your equation is incorrect. This is simply a comma , in the original printed formula. –  pbs Feb 8 '14 at 17:28

Using the tangent addition formulas, we get

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

and

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

From this, we get

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

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

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

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

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

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

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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.

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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 ;)

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