Question

If $7\alpha =2\pi$, then find the absolute value of $sec\alpha +sec2\alpha +sec4\alpha$.

Answer 1

Given, $7\alpha =2\pi$,

The absolute value of,

$\sec \alpha +\sec 2\alpha +\sec 3\alpha$

$=|\sec \alpha +\sec 2\alpha +\sec 3\alpha |$

$=|\sec \frac{2\pi }{7}+\sec 2\frac{2\pi }{7}+\sec 3\frac{2\pi }{7}|$

$=|\frac{1}{\cos \frac{2\pi }{7}}+\frac{1}{\cos 2\frac{2\pi }{7}}+\frac{1}{\cos 3\frac{2\pi }{7}}|$

We have the formula $\cos 2A=2{\cos }^{2}A-1$ and $\cos 4A=1-2(2\sin A\cos A{)}^2$

$=|\frac{1}{\cos \frac{2\pi }{7}}+\frac{1}{2{\cos }^2\frac{2\pi }{7}-1}+\frac{1}{1-2(2\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}{)}^2}|$

$\cos \frac{2\pi }{7}=-0.2225$ and $\sin \frac{2\pi }{7}=0.9749$

Substituting the values,

$=|\frac{1}{-0.2225}+\frac{1}{2(-0.2225{)}^2-1}+\frac{1}{1-2\left(2\right(0.9749\left)\right(-0.2225){)}^2}|$

$=|-4.4943-1.1093+1.6036|=|-4|$

$\therefore$The absolute value of, $\sec \alpha +\sec 2\alpha +\sec 3\alpha$ is $4$