Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$

Question

Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$,
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Also we know that all the expressions are different from zero and defined. Including the expressions we received during the solution.
Tried to play with it, didn't seem to work for me.

They mentioned all the all the expressions are different from zero, probably also $\alpha,\beta$ — TuYu, Dec 08 '17 at 19:41

score 2 · Accepted Answer · answered Dec 08 '17 at 20:10

2

Applying sum to product rule on LHS and sine of the sum of 2 angles on RHS

$$2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$$

$$4\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)=6\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha+\beta}{2}\right)$$

$$2\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=3\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$

$$2\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)+2\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=3\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)-3\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)$$

$$5\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)$$

$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)=\frac{1}{5} \quad \square$$

answered Dec 08 '17 at 20:10

user

154,566

How did you get to the third line? – TuYu Dec 08 '17 at 20:17
cancel out sin or are you referring to the second equality? – user Dec 08 '17 at 20:20
It was what you said, now I understand. Thanks! – TuYu Dec 08 '17 at 20:21
You are welcome! sum to product are very useful! Bye – user Dec 08 '17 at 20:23
@gimusi, After $$2\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=3\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$
we can apply https://brilliant.org/wiki/componendo-and-dividendo/ on $$\dfrac32=\dfrac{\cos\dfrac{\alpha-\beta}2}{\cos\dfrac{\alpha+\beta}2}$$
– lab bhattacharjee Dec 09 '17 at 10:58

Badam Baplan · Answer 2 · 2017-12-08T19:50:41.183

Recall the angle addition formula $$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$ for arbitrary angles.

From what you are given,

$$\frac{2}{3}\sin(\alpha)+\frac{2}{3}\sin(\beta)=\sin(\alpha+\beta)$$

So let's say $\cos(\beta) = \frac{2}{3} = cos(\alpha)$

Now your question is reduced to showing that $$ \frac{1}{\sqrt{5}} = \tan\big(\frac{\cos^{-1}(\frac{2}{3})}{2}\big)$$

Option one, reason directly from a triangle. Option two, you can now use the relevant half-angle formula,

$$\tan(\frac{\alpha}{2}) = \sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}$$

which shows $$\tan(\frac{\alpha}{2}) = \sqrt{\frac{1-2/3}{1+2/3}} = \sqrt{1/5}$$

Technically correct but detour – lab bhattacharjee Dec 08 '17 at 19:44 — lab bhattacharjee, Dec 08 '17 at 19:44

Jean Marie · Answer 3 · 2017-12-08T20:20:19.320

Using the formulas of half-angle substitution, also called Weierstrass substitution formulas,

$$\cos(u)=\dfrac{1-t^2}{1+t^2} , \ \sin(u)=\dfrac{2t}{1+t^2} \ \ \ \text{where} \ \ \ t:=\tan(\tfrac{u}{2}),$$

one has to prove, setting $a=\tan(\alpha/2)$ and $b=\tan(\beta/2)$, that :

$$2\dfrac{2a}{1+a^2}+2\dfrac{2b}{1+b^2}=3 \dfrac{2a}{1+a^2}\dfrac{1-b^2}{1+b^2} +3 \dfrac{2b}{1+b^2}\dfrac{1-a^2}{1+a^2} \implies ab=\dfrac15$$

(we have used $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$).

The first expression, once we have multiplied both sides by $(1+a^2)(1+b^2)$ and cancelled by 2, gives:

$$2a+2ab^2+2b+2ba^2 = 3a-3a^2b+3b-3a^2b$$

$$\iff 5ab(a+b)=a+b$$

giving indeed, under the supplementary assumption that $a+b \neq 0$

$$ab=\tfrac15.$$

Remark: the overall advantage of formulas (*) is that they convert trigonometrical issues into fully algebraic formulations.

Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$

3 Answers3