Question

1+cos(18x)=_____.

a) 2sin^2 (18x)
b) 2cos^2 (9x)
c) 2cos^2 (18x)
d) 2sin^2 (9x)

Answer 1

@jim_thompson5910

Answer 2

@mathstudent55

Answer 3

hi need help

Answer 4

yes i need help

Answer 5

i gave you a medal

Answer 6

with what

Answer 7

\[\Large 1 + \cos(18x) = 1 + \cos(2*9x)\]


\[\Large 1 + \cos(18x) = 1 + 2\cos^2(9x) - 1\]


\[\Large 1 + \cos(18x) = 2\cos^2(9x)\]


On step 2, I used this identity \[\Large \cos(2x) = \cos^2(x) - 1\] see this link for more trig identities
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
specifically see page 2 and the "Double Angle Formulas" section

Answer 8

oops I made a typo, I meant to say \[\Large \cos(2x) = 2\cos^2(x) - 1\]

Answer 9

how do i use that equation to find the answer

Answer 10

@jim_thompson5910

Answer 11

someone please help

Answer 12

I practically gave you the answer. Read through my steps again.

Answer 13

oh duh.

Answer 14

haha

Answer 15

thanks!

Answer 16

You use the identity to replace cos(2*9x) with 2*cos^2(9x) - 1

the "9x" is thought of as the theta in \[\Large \cos(2\theta) = 2\cos^2(\theta) - 1\]

Answer 17

np