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

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.

Quanto
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I_love_maths
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    Can you explain what do you mean by sum and product rule in the question? It's a good thing you are trying to explain your attempt but notice that not all of us may share the same terminology as you. For example, you may call certain ideas with certain names we do not know. Key point is that: Always try to explain your terminology, if you think it's not standard. – Someone May 09 '20 at 05:40
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    $$\cos6=\sin(90+6),\sin12=-\sin(180+12)$$ use https://math.stackexchange.com/questions/1591220/frac1-sin-8-circ-frac1-sin-16-circ-frac1-sin-4096-circ – lab bhattacharjee May 09 '20 at 05:42
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    @mann I meant sin a * sin b and sin a + sin b identities – I_love_maths May 09 '20 at 06:40

2 Answers2

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\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}

Quanto
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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 $n=1$

lab bhattacharjee
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