$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if...........

Question

Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can I approach

Answer 1

Hint

These equalities are useful $$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$$

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

Answer 2

Use the following identities: $$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)\\ \cos\beta-\cos\alpha=-2\sin\left(\frac{\beta+\alpha}2\right)\sin\left(\frac{\beta-\alpha}2\right)$$ So the first condition is equivalent to $$\cos\left(\frac{\alpha-\beta}2\right)=-\sqrt{3}\sin\left(\frac{\beta-\alpha}2\right)$$ I.e $$\tan\left(\frac{\alpha-\beta}2\right)=\frac{1}{\sqrt{3}}$$ Then use that $1+\tan^2\theta=\frac1{\cos^2\theta}$

Answer 3

Apart from the Prosthaphaeresis Formulas already mentioned with the unmentioned assumption that $\displaystyle \sin\frac{\alpha+\beta}2\ne0$

we can try as follows :

Rearranging we have $\displaystyle\sin\alpha+\sqrt3\cos\alpha=\sqrt3\cos\beta-\sin\beta$

As $\displaystyle 1^2+(\sqrt3)^2=4,$ we write $1=2\sin30^\circ,\sqrt3=2\cos30^\circ$ to get

$\displaystyle2\sin30^\circ\sin\alpha+2\cos30^\circ\cos\alpha=2\cos30^\circ\cos\beta-2\sin30^\circ\sin\beta$

$\displaystyle\implies \cos(\alpha-30^\circ)=\cos(30^\circ+\beta)$

$\displaystyle\implies \alpha-30^\circ=n360^\circ\pm(30^\circ+\beta)$

Taking the '-' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ-(30^\circ-\beta)\implies \alpha=n360^\circ-\beta$

This makes both sides equal for all $\alpha,$ so no solution available from here

Taking the '+' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ+(30^\circ-\beta)\implies \alpha-\beta=n360^\circ+60^\circ$

Now, use $\displaystyle2\cos^2A=1+\cos2A$ as $\displaystyle\cos2A=2\cos^2A-1$ for $\displaystyle A=\frac{\alpha-\beta}2$