Anyone knows how to prove this equation (using trigonometric identities?)

Prove: $$\frac {\sin \beta \cos \beta \pm \sin \alpha \cos \alpha}{\cos^2 \beta - \sin^2 \alpha}=\tan(\beta \pm \alpha)$$

For example: $$\beta = 3^\circ$$ And $$\alpha =23^\circ$$, both sides of this equation are $$0.4877325$$ for the $$+$$ case. I also substituted other different values into this equation and they are all equal, but I do not know how to prove it. It's from a book, just said using trigonometric identities, but I googled a lot, still cannot prove it. Anyone can help me to prove it? Thanks!

• Is there any chance you would consider an operation called cropping the image? – String Feb 5 '15 at 19:53
• Better yet - use MathJaX – String Feb 5 '15 at 19:53
• Do you know the sine and cosine angle sum formulas? – user7530 Feb 5 '15 at 19:57
• Any chance that you might make the slightest attempt? – copper.hat Feb 5 '15 at 19:59
• I googled “trigonometric identities” and found the one you need on the first screenful of text on the first linked page. – Matthew Leingang Feb 5 '15 at 20:06

It can done using five standard trig identities: $$(1)\qquad\sin2\theta=2\sin\theta\cos\theta,$$ $$(2\text{a,b})\quad\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta,$$ $$(3)\quad\sin2\theta+\sin2\phi=2\sin(\theta+\phi)\cos(\theta-\phi),$$ $$(4)\quad\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi).$$ Then $$\frac {\sin \beta \cos \beta + \sin \alpha \cos \alpha}{\cos^2 \beta - \sin^2 \alpha}$$ $$=\frac{\sin2\beta+\sin2\alpha}{\cos2\beta+\cos2\alpha}\quad(\text{by identities 1 and 2a,b).}$$ Using identities 3 and 4, and cancelling $\cos(\alpha-\beta)$, now gives the required result for the plus sign. The result for the minus sign is obtained by replacing $\alpha$ by $-\alpha$.

• You are amazing! this is surely correct! Thanks very much!!! I will vote you up as soon as I get enough reputation – Yas Feb 5 '15 at 23:39
• @Yas: Even with minimal rep, a questioner can award points by accepting an answer. There might be a better answer on its way, so I suggest waiting a while to see what answers are posted before choosing one to accept. – John Bentin Feb 5 '15 at 23:57
• Thanks for letting me know this, this is my first day on StackExchange, a lot of stuff need to learn. Thanks again for your answer~ :) – Yas Feb 6 '15 at 0:37

If you are willing to start from the identity for the tangent of a sum $$\tan(\theta \pm \gamma) = \frac{\tan\theta \pm \tan \gamma}{1\mp \tan\theta \tan \gamma}$$ then write each $\tan x$ as $\frac{\sin x}{\cos x}$ and then multiply the numerator and denominator by $\cos \theta \cos \gamma$ to get the identity you are trying to prove.

Or you can use the sum of angles formulas for sine and cosine $$\sin (\theta \pm \gamma) = \sin \theta \cos \gamma \pm \cos \theta \sin \gamma \\ \cos (\theta \pm \gamma) = \cos\theta\cos\gamma \mp \sin\theta\sin\gamma$$ and combine them to form $\tan(\theta\pm\gamma)$ getting the same result.

• It seems to me that in the answer is not proved the formula from the question. Is it? – Janko Bracic Feb 5 '15 at 20:14
• Thanks for suggestion. But it does not prove it, because the term in my numerator is $\sin \beta \cos \beta$, not $\sin \beta \cos \alpha$ – Yas Feb 5 '15 at 20:37