Question

convert into a product and simplify
cos(5pi/12+theta)+cos(pi/12-theta)

Answer 1

i got the answer
\[-\frac{ \sqrt{2} }{ 2 }\sin (\theta)\]

Answer 2

is that right?

Answer 3

\[\frac{ \sqrt{2} }{ 2 }\sin (-\theta)\]

Answer 4

\[\cos (\frac{ 5\pi }{ 12 }+\theta)\cos(\frac{ \pi }{ 12 }-\theta)\]
original question

Answer 5

is it

\[\cos (\frac{ 5\pi }{ 12 }+\theta)\cos(\frac{ \pi }{ 12 }-\theta)\]

or

\[\cos (\frac{ 5\pi }{ 12 }+\theta)+\cos(\frac{ \pi }{ 12 }-\theta)\]

Answer 6

the second one.. sorry. mistype

Answer 7

I'm going to use the identity

\[\Large \cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\]

Answer 8

isnt it -2cos(A+B/2).... not sure if it matters.. just what our book lists.

Answer 9

that looks like part of it

Answer 10

i used the wrong identity then.. ugh.

Answer 11

yeah the identity i wrote is correct because it is listed here

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

Answer 12

in this case,

A = 5pi/12 + theta

B = pi/12 - theta


So

A+B = (5pi/12 + theta) + (pi/12 - theta)

A+B = pi/2

(A+B)/2 = pi/4

-----------------------------------------

A-B = (5pi/12 + theta) - (pi/12 - theta)

A-B = pi/3 + 2theta

(A-B)/2 = pi/6 + theta

Answer 13

so \[\frac{ \sqrt{6} }{ 2 }\cos(\theta)\]

Answer 14

cos((A+B)/2) = cos(pi/4) = sqrt(2)/2

cos((A-B)/2) = cos(pi/6 + theta)


So put this all together to get


\[\Large \cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\]

\[\Large \cos\left(\frac{ 5\pi }{ 12 }+\theta\right) + \cos\left(\frac{ \pi }{ 12 }-\theta\right) = 2\left(\frac{\sqrt{2}}{2}\right)\left(\cos\left(\frac{\pi}{6} + \theta\right)\right)\]


\[\Large \cos\left(\frac{ 5\pi }{ 12 }+\theta\right) + \cos\left(\frac{ \pi }{ 12 }-\theta\right) = \sqrt{2}\cos\left(\frac{\pi}{6} + \theta\right)\]

Answer 15

so you dont convert the cos(pi/6) to sqrt(3)/2?

Answer 16

no you can't do that because it's not just cos(pi/6)

it's cos(pi/6 + theta)

Answer 17

if it were just cos(pi/6), then yes, you can do that

Answer 18

so just distribute the 2 to cos(pi/4) and leave the 2nd cosine term. right? just making sure.. have a quiz in an hour and a half :)

Answer 19

yes because cos(pi/4) is really sqrt(2)/2

Answer 20

perfect. thanks for your help once again. I understand now.

Answer 21

yw