I need help with this question

[Thank You Vice President]

I understood how to do numbers 1 and 2. I dont know how to do number 3. How would I do that? Thanks

 

[Dr.Peterson]

I see no 1, 2, 3 here. Do you mean #18?

Please show how you did #16, and I'll show you how to use the same method to do #18. (There may also be an easier way, in which case I'll show you that, too.)

 

[Thank You Vice President]

I found out how to do it, it was an even and odd function. Cosine(-theta)=cosine(+theta), so its -15/22

 

[Steven G]

I jump the gun and state that cos(-$\theta$) = cos($\theta$) implying that the cos function is even.

 

[R.M.]

What is the cos(30)? What is the cos(-30)? How about cos(60) and cos(-60)? See if a pattern begins to emerge.

 

[pka]

What is the cos(30)? What is the cos(-30)? How about cos(60) and cos(-60)?
See if a pattern begins to emerge.

Because cosine is an even function $\cos(x)=\cos(-x)$.
$\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt3}{2}$ and
$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$

 

[Dr.Peterson]

I found out how to do it, it was an even and odd function. Cosine(-theta)=cosine(+theta), so its -15/22

Yes, that was the "easier way" I would have told you if your method had been to use angle subtraction identities!