# cos32 + cos58 + cos20 + cos70

• Feb 19th 2006, 08:13 PM
unit
cos32 + cos58 + cos20 + cos70
I ran into this problem on a worksheet:

Simplify and Evaluate: (without using a calculator)
cos32 + cos58 + cos20 + cos70

I have tried using difference of sine and difference of cosine by
cos (30 +2) = cos 30cos2 - sin30sin2
etc.. with the others

still nothing comes to mind

anyone want to give it a shot?
• Feb 19th 2006, 08:43 PM
earboth
Quote:

Originally Posted by unit
I ran into this problem on a worksheet:

Simplify and Evaluate: (without using a calculator)
cos32 + cos58 + cos20 + cos70

I have tried using difference of sine and difference of cosine by
cos (30 +2) = cos 30cos2 - sin30sin2
etc.. with the others
still nothing comes to mind

anyone want to give it a shot?

Hello,

1. use $\displaystyle \cos(\alpha)+\cos(\beta)=2\cos \frac{\alpha+\beta}{2}\ \cdot \ 2\cos \frac{\alpha-\beta}{2}$

2. The first 2 angles add up to 90°, the 2nd two too. So the first factor on the RHS of the equation will be $\displaystyle \sqrt{2}$

I hope this will help you a little bit further on. And now I've to hurry: Duty is calling!

Greetings

EB
• Feb 20th 2006, 04:06 PM
unit
Are there any other ways of doing this problem?

I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

Maybe you could show me how to derive this? All help greatly appreciated! Thanks!
• Feb 21st 2006, 02:42 PM
ThePerfectHacker
Quote:

Originally Posted by unit
Are there any other ways of doing this problem?

I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

Maybe you could show me how to derive this? All help greatly appreciated! Thanks!

Notice that,
$\displaystyle 2\cos x\cos y=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y$
Thus, using the formula:
$\displaystyle \cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$
Thus, we have that:
$\displaystyle 2\cos x\cos y=\cos(x+y)+\cos(x-y)$
Now let,
$\displaystyle A=x+y$
$\displaystyle B=x-y$
Thus,
$\displaystyle x=\frac{A+B}{2}$
$\displaystyle y=\frac{A-B}{2}$
Substituting this we get,
$\displaystyle \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
Q.E.D.