If $$\cos\theta = \frac{\cos\alpha + \cos \beta}{1 + \cos\alpha\cos\beta}$$ then prove that one of the values of $\tan{\frac{\theta}{2}}$ is $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$.
I don't even know how to start this question. Pls help
Answer
I think i have found out how to approach it. I hope this proof is satifactory.
Using the half angle formula,
$$\tan{\frac{\theta}{2}} = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \longrightarrow \text{eq.1}$$
Evaluating $\frac{1 - \cos\theta}{1 + \cos\theta}$ first,
$$\frac{1 - \cos\theta}{1 + \cos\theta} = \frac{1 - (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}{1 + (\frac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta})}\text{ }[\text{since } \cos\theta= \frac{\cos\alpha +\cos\beta}{1 + \cos\alpha\cos\beta}]$$
$$=\frac{1+\cos\alpha\cos\beta - \cos\alpha -\cos\beta}{1+\cos\alpha\cos\beta + \cos\alpha+ \cos\beta}$$
$$=\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}$$
Substituting this value of $\frac{1 - \cos\theta}{1 + \cos\theta}$ into equation 1,
$$\tan{\frac{\theta}{2}} = \sqrt{\frac{(1-\cos\alpha)(1-\cos\beta)}{(1+\cos\alpha)(1+\cos\beta)}}$$
$$=\sqrt{\frac{(1-\cos\alpha)^2(1-\cos\beta)^2}{(1-\cos^2\alpha)(1-\cos^2\beta)}}$$
$$=\pm\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta}$$
Taking the positive value of $\tan{\frac{\theta}{2}}$,
$$\frac{(1-\cos\alpha)(1-\cos\beta)}{\sin\alpha\sin\beta} = \frac{4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}}{4\sin\frac{\alpha}{2}\cos{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}} \text{(using half angle and double angle formula)}$$
$$=\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$$
Therefore, $\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}$ is one of the values of $\tan{\frac{\theta}{2}}$.
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