MHB How to Prove this Trigonometric Identity?

[anemone]

Prove $$\frac{\sin\left(5\tfrac{3}{4}^{\circ} \right)}{\cos\left(17\tfrac{1}{4}^{\circ} \right)}+\frac{\sin\left(17\tfrac{1}{4}^{\circ} \right)}{\cos\left(51\tfrac{3}{4}^{\circ} \right)}+\frac{\sin\left(51\tfrac{3}{4}^{\circ} \right)}{\cos\left(155\tfrac{1}{4}^{\circ} \right)}=\frac{1}{2}\left(\tan\left(155\tfrac{1}{4}^{\circ} \right)-\tan\left(5\tfrac{3}{4}^{\circ} \right) \right)$$

 

[Opalg]

Prove $$\frac{\sin\left(5\tfrac{3}{4}^{\circ} \right)}{\cos\left(17\tfrac{1}{4}^{\circ} \right)}+\frac{\sin\left(17\tfrac{1}{4}^{\circ} \right)}{\cos\left(51\tfrac{3}{4}^{\circ} \right)}+\frac{\sin\left(51\tfrac{3}{4}^{\circ} \right)}{\cos\left(155\tfrac{1}{4}^{\circ} \right)}=\frac{1}{2}\left(\tan\left(155\tfrac{1}{4}^{\circ} \right)-\tan\left(5\tfrac{3}{4}^{\circ} \right) \right)$$

Let $\theta = 5{\frac34}^\circ$. Then the problem becomes

Prove $$\frac{\sin\theta}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} = \tfrac12\bigl(\tan(27\theta) - \tan\theta\bigr)$$.

Suppose we knew that $$\frac{\sin x}{\cos(3x)} = \tfrac12\bigl(\tan(3x) - \tan x\bigr).\quad(*)$$

Putting $x$ equal to $\theta$, then $3\theta$, and then $9\theta$, it would follow that $$\begin{aligned}\frac{\sin\theta}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} &= \tfrac12\bigl(\tan(3\theta) - \tan\theta\bigr)+ \tfrac12\bigl(\tan(9\theta) - \tan(3\theta)\bigr)+ \tfrac12\bigl(\tan(27\theta) - \tan(9\theta)\bigr) \\ &= \tfrac12\bigl(\tan(27\theta) - \tan\theta\bigr), \end{aligned}$$ as required. It just remains to prove the identity (*), which I'll leave for someone else.

 

[kaliprasad]

Let $\theta = 5{\frac34}^\circ$. Then the problem becomes

Suppose we knew that $$\frac{\sin x}{\cos(3x)} = \tfrac12\bigl(\tan(3x) - \tan x\bigr).\quad(*)$$

Putting $x$ equal to $\theta$, then $3\theta$, and then $9\theta$, it would follow that $$\begin{aligned}\frac{\sin\theta}{\cos(3\theta)} + \frac{\sin(3\theta)}{\cos(9\theta)} + \frac{\sin(9\theta)}{\cos(27\theta)} &= \tfrac12\bigl(\tan(3\theta) - \tan\theta\bigr)+ \tfrac12\bigl(\tan(9\theta) - \tan(3\theta)\bigr)+ \tfrac12\bigl(\tan(27\theta) - \tan(9\theta)\bigr) \\ &= \tfrac12\bigl(\tan(27\theta) - \tan\theta\bigr), \end{aligned}$$ as required. It just remains to prove the identity (*), which I'll leave for someone else.

we have tan 3x - tan x = sin 3x/ cos 3x - sin x/ cos x
= ( sin 3x cos x - cos 3x sin x)/ ( cos 3x sin x)
= sin 2x/( cos3x sin x)
= ( 2 sin x cos x)/(cos 3x sin x)
= 2 sin x / cos 3x
or sin x/ cos 3x = 1/2( tan 3x - tan x)