Prove $\frac1{\cos6°}+ \frac1{\sin24°} + \frac1{\sin48° }- \frac1{\sin12° }= 0$

Question

Prove that $$\frac1{\cos6°}+ \frac1{\sin24°} + \frac1{\sin48° }- \frac1{\sin12° }= 0$$

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I tried to solve this question by grouping together the first and last term, and applying the sum and product rule, (same for the middle two terms). But I wasn't able to do it that way.

Any help?

PS: I am just a 15 year old, so please be patient and explain line by line (preferably with only trig identities). Thanks.

Answer 1

\begin{align} & \frac1{\cos6} + \frac1{\sin24} + \frac1{\sin48} - \frac1{\sin12} \\ = &\left(\frac1{\cos6} + \frac1{\sin48}\right) + \left(\frac1{\sin24}- \frac1{\sin12}\right) \\ = &\frac{\sin48+\sin96}{\cos6\sin48}+\frac{\sin12-\sin24}{\sin12\sin24} \\ =&\frac{2\sin72\cos24}{2\cos6\cos24\sin24}-\frac{2\cos18\sin6}{2\cos6\sin6\sin24} \\ = &\frac{\cos18}{\cos6\sin24}-\frac{\cos18}{\cos6\sin24} =0\\ \end{align}

Answer 2

Let us see how the problem came into being:

Like $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$

for $\sin2A\ne0,$ using $$\cot A-\cot2A=\dfrac1{\sin2A}$$

if $\sin16x\ne0,$ $$f(x)=\dfrac1{\sin2x}+\dfrac1{\sin4x}+\dfrac1{\sin8x}+\dfrac1{\sin16x}=\cot x-\cot16x$$

Now $f(x)=0$ if $\cot16x=\cot x,16x=180^\circ n+x\iff x=12^\circ n$ where $n$ is any integer

Here $n=1$