# Thread: An awkward cosine problem

1. ## An awkward cosine problem

A student came to me with this problem, and sadly I do not know how to finish this problem, any help, please?

$\cos ^2 ( \frac { \pi }{8} - \frac {1}{2} )$

2. What exactly do you want to do with this?

A student came to me with this problem, and sadly I do not know how to finish this problem, any help, please?

$\cos ^2 ( \frac { \pi }{8} - \frac {1}{2} )$
What is the problem, to evaluate that? Use your claculator, it is very unlikely to have a simple closed form.

You can mess about with it using trig identities to expand the cos and then simplifying, so:

$\cos^2\left(\frac{\pi/4-1}{2}\right)=\frac{1+\cos(\pi/4-1)}{2}$

Now expand the $\cos$ on the right ..

CB

4. Umm... I'm still having some difficulties in expanding the right cos...

I have $\cos ( \frac { \pi } {4} - 1 ) = \frac { \sqrt {2} }{2} cos (1) + \frac { \sqrt {2} }{2} sin (1)$

So I should have $\frac { \sqrt {2} }{4} \cos (1) + \frac { \sqrt {2} }{4} \sin (1) + \frac {1}{2}$

but then how do I simplify that? The answer should have been $\frac { \pi }{4}$ according to the book...

Umm... I'm still having some difficulties in expanding the right cos...

I have $\cos ( \frac { \pi } {4} - 1 ) = \frac { \sqrt {2} }{2} cos (1) + \frac { \sqrt {2} }{2} sin (1)$

So I should have $\frac { \sqrt {2} }{4} \cos (1) + \frac { \sqrt {2} }{4} \sin (1) + \frac {1}{2}$

but then how do I simplify that? The answer should have been $\frac { \pi }{4}$ according to the book...
In that case either you have given us the wrong question or the book is wrong

$(\cos(\frac{\pi}{8}-\frac{1}{2}))^2 \approx 0.9885$

while $\pi/4 \approx 0.7854$

CB