Question

Rewrite with only sin x and cos x.

cos 2x - sin x

Answer 1

@bohotness

Answer 2

@mathmate

Answer 3

2x - sin x, how to rewrite this any helpful notes

Answer 4

what??

Answer 5

Use the double angle formulae:
\(\rm sin(2x)=2sin(x)cos(x)\)
\(\rm cos(2x)=cos^2(x)-sin^2(x)\)

Answer 6

(1) First, find sin(3x) and cos(3x) in terms of sine and cosine as follows:
sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x), by the sine addition formula
= 2sin(x)cos^2(x) + sin(x)[1 - 2sin^2(x)], by the sine/cosine double-angle formulas
= 2sin(x)[1 - sin^2(x)] + sin(x)[1 - 2sin^2(x)], by the Pythagorean Identity
= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)
= 3sin(x) - 4sin^3(x).

cos(3x) = cos(2x + x)
= cos(2x)cos(x) - sin(2x)sin(x), by the cosine addition formula
= cos(x)[2cos^2(x) - 1] - 2sin^2(x)cos(x), by the sine/cosine double-angle formulas
= cos(x)[2cos^2(x) - 1] - 2cos(x)[1 - cos^2(x)], by the Pythagorean Identity
= 2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
= 4cos^3(x) - 3cos(x).

Thus:
sin(3x) + cos(3x) = [3sin(x) - 4sin^3(x)] + [4cos^3(x) - 3cos(x)]
= 4cos^3(x) - 4sin^3(x) + 3sin(x) - 3cos(x).

(2) By the double-angle formula for cosine:
cos(2x) - sin(x) = 1 - 2sin^2(x) - sin(x).

Answer 7

@jim_thompson5910

Answer 8

so the answer is cos^2x-sin^3x?

Answer 9

:DDDDDDDDDDDDDDDDD

Answer 10

@bohotness
I don't quite get where you're going with cos(3x), etc.
But the second part makes sense!

Answer 11

heres my choice
1) cos^2x - sin^2x - sinx
2) cos^2x - sin^3x
3) cos^2x + sin^2x
4) cos^2x - 3sin x

Answer 12

ybarrap answered this already

http://openstudy.com/users/dglicks#/updates/54f506b0e4b09f47f4e49152

Answer 13

didnt expect him to just gimme the answer like that lol thanks jim