Tan of difference of two angles given as sum of sines and cosines

Question

I could have done the question had it been to find the value of $\tan\frac{\alpha+\beta}{2}$. But here the tangent of the difference has been asked. How does one go about it?

If $\sin\alpha + \sin\beta = a$ and $\cos\alpha + \cos\beta = b$, then show that $$\tan\frac{\alpha-\beta}{2}=\pm\sqrt{\frac{4-a^2-b^2}{a^2 + b^2}}.$$

Answer 1

From the half-angle formula for $\tan$, we have $$\tan\frac{\alpha-\beta}2=\pm\sqrt{\frac{1-\cos(\alpha-\beta)}{1+\cos(\alpha-\beta)}}\tag{1}$$

Now $$a^2=(\sin\alpha+\sin\beta)^2=\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta$$ $$b^2=(\cos\alpha+\cos\beta)^2=\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$$ so

$$a^2 + b^2=2+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=2(1+\cos(\alpha-\beta))$$ Hence $$\pm\sqrt{\frac{4-a^2-b^2}{a^2 + b^2}}=\pm\sqrt{\frac{4-2(1+\cos(\alpha-\beta))}{2(1+\cos(\alpha-\beta))}}=\pm\sqrt{\frac{1-\cos(\alpha-\beta)}{1+\cos(\alpha-\beta)}}$$ which is the same as $(1)$.

Answer 2

Use $$a^2+b^2=2+\cos(\alpha-\beta).$$ We obtain: $$\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}=\sqrt{\frac{2-2\cos(\alpha-\beta)}{2+2\cos(\alpha-\beta)}}=\left|\tan\frac{\alpha-\beta}{2}\right|$$

Answer 3

For $0 \leq \beta \leq \alpha \leq 90^\circ$ ...

$$\tan\frac{\alpha-\beta}{2} = \frac{d}{c} = \sqrt{\frac{2^2-c^2}{c^2}} = \sqrt{\frac{4-\left(a^2+b^2\right)}{a^2+b^2}}$$