MHB Proving a trigonometric identity

[maxkor]

How prove $\cos\frac{8\pi}{35}+\cos\frac{12\pi}{35}+\cos\frac{18\pi}{35}=\frac{1}{2}\cdot\left(\cos\frac{\pi}{5}+\sqrt7\cdot\sin\frac{\pi}{5}\right)$?

 

[MarkFL]

I have retitled the thread, since a title of "trig" in our Trigonometry forum tells our readers no more that they would already surmise. A good thread title briefly describes the question being asked.

Can you post what you have tried so far so our helpers know where you are stuck, and won't offer suggestions that you may have already tried?

 

[maxkor]

$\cos \frac{12\pi}{35}=\cos( \frac{\pi}{5}+ \frac{\pi}{7})=\cos \frac{\pi}{5} \cdot \cos \frac{\pi}{7}-\sin \frac{\pi}{5} \cdot \sin \frac{\pi}{7}$$\cos \frac{8\pi}{35}=\cos( -\frac{\pi}{5}+ \frac{3\pi}{7})=\cos \frac{\pi}{5} \cdot \cos \frac{3\pi}{7}+\sin \frac{\pi}{5} \cdot \sin \frac{3\pi}{7}$$\cos \frac{18\pi}{35}=-\cos( \frac{\pi}{5}+ \frac{2\pi}{7})=-\cos \frac{\pi}{5} \cdot \cos \frac{2\pi}{7}+\sin \frac{\pi}{5} \cdot \sin \frac{2\pi}{7}$

what next?

 

[MarkFL]

I believe you want instead:

$$\cos\left(\frac{18\pi}{35}\right)=\cos\left(-\frac{\pi}{5}+\frac{5\pi}{7}\right)$$

Once you expand that like your first two equations, then add and factor on the two trig. expressions on the right side of the identity you are given to prove. Then you will have two identities resulting from equating the coefficients you must prove.

 

[MarkFL]

I see now I missed the negative sign, and indeed:

$$\cos\left(\frac{18\pi}{35}\right)=-\cos\left(\frac{17\pi}{35}\right)$$

So, add what you have, and factor as I suggested above. :D