Prove the following identities :
Exercise Problems & Solutions
Prove that: $\sin10^{\circ}\sin 50^{\circ} + \sin50^{\circ}\sin250^{\circ} + \sin250^{\circ}\sin10^{\circ} = -\frac{3}{4}$
Solution:
\[ = \frac{1}{2}(2\sin10^{\circ}\sin50^{\circ}) + \frac{1}{2}(2\sin50^{\circ}\sin250^{\circ}) + \frac{1}{2}(2\sin250^{\circ}\sin10^{\circ}) \] \[ = \frac{1}{2}\left[\cos(50^{\circ}-10^{\circ}) - \cos(50^{\circ}+10^{\circ}) + \cos(250^{\circ}-50^{\circ}) - \cos(250^{\circ}+50^{\circ}) + \cos(250^{\circ}-10^{\circ}) - \cos(250^{\circ}+10^{\circ})\right] \] \[ = \frac{1}{2}\left[\cos40^{\circ} - \cos60^{\circ} + \cos200^{\circ} - \cos300^{\circ} + \cos240^{\circ} - \cos260^{\circ}\right] \] \[ = \frac{1}{2}\left[\cos40^{\circ} - \frac{1}{2} + \cos(2\times90^{\circ}+20^{\circ}) - \cos(4\times90^{\circ}-60^{\circ}) + \cos(3\times90^{\circ}-30^{\circ}) - \cos(3\times90^{\circ}-10^{\circ})\right] \] \[ = \frac{1}{2}\left[\cos40^{\circ} - \frac{1}{2} - \cos20^{\circ} - \cos60^{\circ} - \sin30^{\circ} + \sin10^{\circ}\right] \] \[ = \frac{1}{2}\left(\cos40^{\circ} - \frac{1}{2} - \cos20^{\circ} - \frac{1}{2} - \frac{1}{2} + \cos(90^{\circ}-10^{\circ})\right) \] \[ = \frac{1}{2}\left(\cos40^{\circ} + \cos80^{\circ} - \cos20^{\circ} - \frac{3}{2}\right) \] \[ = \frac{1}{2}\left(\sin(90^{\circ}-40^{\circ}) + 2\sin\frac{80^{\circ}+20^{\circ}}{2}\sin\frac{20^{\circ}-80^{\circ}}{2} - \frac{3}{2}\right) \] \[ = \frac{1}{2}\left(\sin50^{\circ} + 2\sin50^{\circ}\sin(-30^{\circ}) - \frac{3}{2}\right) \] \[ = \frac{1}{2}\left(\sin50^{\circ} + 2\sin50^{\circ}\left(-\frac{1}{2}\right) - \frac{3}{2}\right) \] \[ = \frac{1}{2}\left(\sin50^{\circ} - \sin50^{\circ} - \frac{3}{2}\right) = -\frac{3}{4} \quad (\text{proved}) \]Prove that: $\frac{\sin\alpha\sin11\alpha + \sin3\alpha\sin7\alpha}{\sin\alpha\cos11\alpha + \sin3\alpha\cos7\alpha} = \tan8\alpha$
Solution:
\[ = \frac{(2\sin\alpha\sin11\alpha) + (2\sin3\alpha\sin7\alpha)}{(2\sin\alpha\cos11\alpha) + (2\sin3\alpha\cos7\alpha)} \] \[ = \frac{\cos(11\alpha-\alpha) - \cos(11\alpha+\alpha) + \cos(7\alpha-3\alpha) - \cos(7\alpha+3\alpha)}{\sin(11\alpha+\alpha) - \sin(11\alpha-\alpha) + \sin(7\alpha+3\alpha) - \sin(7\alpha-3\alpha)} \] \[ = \frac{\cos10\alpha - \cos12\alpha + \cos4\alpha - \cos10\alpha}{\sin12\alpha - \sin10\alpha + \sin10\alpha - \sin4\alpha} \] \[ = \frac{\cos4\alpha - \cos12\alpha}{\sin12\alpha - \sin4\alpha} = \frac{2\sin\frac{12\alpha+4\alpha}{2}\sin\frac{12\alpha-4\alpha}{2}}{2\sin\frac{12\alpha-4\alpha}{2}\cos\frac{12\alpha+4\alpha}{2}} = \tan8\alpha \quad (\text{proved}) \]Prove that: $\tan\theta\tan\left(\frac{\pi}{3}+\theta\right)\tan\left(\frac{\pi}{3}-\theta\right) = \tan3\theta$
Solution:
\[ = \tan\theta \times \frac{\sin\left(\frac{\pi}{3}+\theta\right)\sin\left(\frac{\pi}{3}-\theta\right)}{\cos\left(\frac{\pi}{3}+\theta\right)\cos\left(\frac{\pi}{3}-\theta\right)} = \tan\theta \times \frac{2\sin\left(\frac{\pi}{3}+\theta\right)\sin\left(\frac{\pi}{3}-\theta\right)}{2\cos\left(\frac{\pi}{3}+\theta\right)\cos\left(\frac{\pi}{3}-\theta\right)} \] \[ = \tan\theta \times \frac{\cos2\theta - \cos120^{\circ}}{\cos120^{\circ} + \cos2\theta} = \tan\theta \times \frac{\cos2\theta + \frac{1}{2}}{\cos2\theta - \frac{1}{2}} = \tan\theta \times \frac{2\cos2\theta + 1}{2\cos2\theta - 1} \] \[ = \frac{\sin\theta}{\cos\theta} \times \frac{2\cos2\theta + 1}{2\cos2\theta - 1} = \frac{2\sin\theta\cos2\theta + \sin\theta}{2\cos\theta\cos2\theta - \cos\theta} = \frac{\sin3\theta - \sin\theta + \sin\theta}{\cos3\theta + \cos\theta - \cos\theta} = \tan3\theta \quad (\text{proved}) \]If $a\cos\phi = b\cos\theta$, show that $a\tan\theta + b\tan\phi = (a+b)\tan\frac{\theta+\phi}{2}$.
Solution:
\[ a\tan\theta + b\tan\phi = \frac{\left(\frac{b\cos\theta}{\cos\phi}\right)\frac{\sin\theta}{\cos\theta} + b\frac{\sin\phi}{\cos\phi}}{\left(\frac{b\cos\theta}{\cos\phi}\right) + b} = \frac{b\sin\theta + b\sin\phi}{b\cos\theta + b\cos\phi} \] \[ = \frac{2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}}{2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}} = \tan\frac{\theta+\phi}{2} \implies a\tan\theta + b\tan\phi = (a+b)\tan\frac{\theta+\phi}{2} \quad (\text{proved}) \]If $p\sin\alpha = q\sin(120^{\circ}+\alpha) = r\sin(240^{\circ}+\alpha)$, prove that $pq + qr + rp = 0$.
Solution:
Let $p\sin\alpha = q\sin(120^{\circ}+\alpha) = r\sin(240^{\circ}+\alpha) = k \quad (k \neq 0)$
\[ \frac{k}{p} + \frac{k}{q} + \frac{k}{r} = \sin\alpha + \sin(120^{\circ}+\alpha) + \sin(240^{\circ}+\alpha) \] \[ = \sin\alpha + 2\sin(180^{\circ}+\alpha)\cos60^{\circ} = \sin\alpha - 2\sin\alpha\left(\frac{1}{2}\right) = \sin\alpha - \sin\alpha = 0 \]Note: $\sin(120^{\circ}+\alpha) + \sin(240^{\circ}+\alpha) = 2\sin(180^{\circ}+\alpha)\cos60^{\circ}$
If $\cos\theta = n\cos(\theta+2\phi)$, show that $(n+1)\tan(\theta+\phi) = (n-1)\cot\phi$.
Solution:
We have, $\cos\theta = n\cos(\theta+2\phi) \quad \text{--- (1)}$
\[ \frac{n+1}{n-1} = \frac{\frac{\cos\theta}{\cos(\theta+2\phi)} + 1}{\frac{\cos\theta}{\cos(\theta+2\phi)} - 1} = \frac{\cos\theta + \cos(\theta+2\phi)}{\cos\theta - \cos(\theta+2\phi)} \] \[ = \frac{2\cos(\theta+\phi)\cos\phi}{2\sin(\theta+\phi)\sin\phi} = \cot(\theta+\phi)\cot\phi \implies (n+1)\tan(\theta+\phi) = (n-1)\cot\phi \quad (\text{proved}) \]Prove that: $\sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha+\beta+\gamma) = 4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2}$
Solution:
\[ = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} + 2\cos\frac{\alpha+\beta+2\gamma}{2}\sin\left(-\frac{\alpha+\beta}{2}\right) \] \[ = 2\sin\frac{\alpha+\beta}{2} \left(\cos\frac{\alpha-\beta}{2} - \cos\frac{\alpha+\beta+2\gamma}{2}\right) \] \[ = 4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2} \quad (\text{proved}) \]Note: $\cos C - \cos D = 2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$
Express $4\sin A \cos B\cos C$ as the sum of four sines.
Solution:
\[ 4\sin A \cos B\cos C = 2(2\sin A \cos B)\cos C \] \[ = 2\left[\sin(A+B) + \sin(A-B)\right]\cos C \] \[ = 2\sin(A+B)\cos C + 2\sin(A-B)\cos C \] \[ = \sin(A+B+C) + \sin(A+B-C) + \sin(A-B+C) + \sin(A-B-C) \]Express $\cos\alpha + \cos\beta + \cos\gamma + \cos(\alpha+\beta+\gamma)$ as the product of three cosines.
Solution:
\[ = (\cos\alpha + \cos\beta) + \left[\cos\gamma + \cos(\alpha+\beta+\gamma)\right] \] \[ = 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} + 2\cos\frac{\alpha+\beta+2\gamma}{2}\cos\frac{\alpha+\beta}{2} \] \[ = 2\cos\frac{\alpha+\beta}{2}\left[\cos\frac{\alpha-\beta}{2} + \cos\frac{\alpha+\beta+2\gamma}{2}\right] \] \[ = 2\cos\frac{\alpha+\beta}{2} \cdot 2\cos\frac{\alpha+\gamma}{2}\cos\frac{\beta+\gamma}{2} \] \[ = 4\cos\frac{\alpha+\beta}{2}\cos\frac{\beta+\gamma}{2}\cos\frac{\gamma+\alpha}{2} \]
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