Question

Rewrite with only sin x and cos x

cos 3x

Answer 1

the choices
cos x - 4 cos x sin2x

-sin3x + 2 sin x cos x

-sin2x + 2 sin x cos x

2 sin2x cos x - 2 sin x cos x

Answer 2

i thought it was the third one

Answer 3

please help i don't know if im right

Answer 4

\(\cos 3x=\cos\left(2x+x\right)\)

Answer 5

\[\cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta\]

Answer 6

i really bad at math i don't understand was my answer choice wrong

Answer 7

i don't know, i didn't work it out. that's your job

Answer 8

can you explain how to do it

Answer 9

use the formula above... then see where it takes you

Answer 10

i got cos x - 4 cos x sin2x thanks

Answer 11

wait i think i did it wrong i re-did it i got -sin3x + 2 sin x cos x

Answer 12

know im just confused

Answer 13

*now*

Answer 14

someone please help me

Answer 15

pgpilot326?

Answer 16

not gettting your choices... could be using difference formula: cos(3x) = cos(4x-x)

Answer 17

umm idk what r u getting

Answer 18

wait...
\[\cos \left( 3x \right)=\cos \left( 2x+x \right)=\cos \left( 2x \right)\cos \left( x \right)-\sin \left( 2x \right)\sin \left( x \right)\]\[=\left[ 1-2\sin^2\left( x\right)\right]\cos \left( x\right) -\left[2\sin \left( x \right) \cos\left( x\right)\right]\sin\left( x \right)\]\[=\cos \left( x \right)-4\sin^2\left( x \right)\cos \left( x \right)\]

Answer 19

its almost like the 2nd answer i got..istill dont get it

Answer 20

it's the first choice if what you mean by sin2x is \(\sin^2 x\)