Question

g(x)=cos^2(pi/8 x)+ sin (pi/4 x), then g(6)

Answer 1

what are you trying to solve for exactly?

Answer 2

g(6) ud just plug in 6 for x right?

Answer 3

Yea just plug it in

Answer 4

\[g(6)=\cos^2(\frac{ 6\pi }{ 8 })+\sin(\frac{ 6\pi }{ 4 })\]

Answer 5

are you trying to simply this out any further?

Answer 6

Yes

Answer 7

ill have to look up the identities

Answer 8

which identities have you recently gone over?

Answer 9

No we haven't gone over any identities yet, this is just a problem from my summer calculus packet.

Answer 10

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 11

you can do all kinds of things to it but

Answer 12

unless your trying to find the reference angle im not sure what exactly we should do

Answer 13

oh

Answer 14

All I know you plug in the given but I don't know how the answer is suppose to look

Answer 15

\[\cos^2(\frac{ 6\pi }{ 8 })+\sin(\frac{ 6\pi }{ 4 })= ?\]

Answer 16

That's right, continue.

Answer 17

well it looks like

Answer 18

the theta in the cos^2 is half the size of the sin theta

Answer 19

Try this: \(\Large \frac{1}{2}(1+cos~2(\frac{3\pi}{4}))+sin(\frac{3\pi}{2})\)

Simplier now?

Answer 20

It'll reduce to: \(\Large \frac{1}{2}(1+0)+(-1)\)

Answer 21

Make sense?

Answer 22

you used product to sum?

Answer 23

Yea I think I get it

Answer 24

I used the half angle formulas: \(\Large \frac{1}{2}(1+cos(2u))\)

I find it easier to factor out the 1/2 right away

Answer 25

^ That one being for \(cos^2x\)

Answer 26

do you have all the identities memorized?

Answer 27

Haha, not all of them, but with practice some just stick to you. Like the half angles are important later on.