Question

wanna make sure I am write on this question
Rewrite with only sin x and cos x.

sin 3x
I got 2 cos2x sin x + sin x - 2 sin3x

Answer 1

\[\sin 3x = 3\sin x - 4\sin^3x\]

Answer 2

?

Answer 3

this is all my choices

Answer 4

2 sin x cos2x + cos x

2 sin x cos2x + sin3x

sin x cos2x - sin3x + cos3x

2 cos2x sin x + sin x - 2 sin3x

Answer 5

sin 3x = sin (2x+x) = sin 2x cos x+ cos2x sinx
= 2sinx cos x cos x+ cos2x sinx
\[= 2sinx \cos ^2x+ (1-\sin^2x) sinx\]
\[= 2sos ^2x inx c+ sinx-\sin^2x sinx \]
\[= 2\cos ^2x sinx + sinx-\sin^3x \]

Answer 6

@llkramer

Answer 7

so I was right?

Answer 8

@llkramer how have you got 2 sin3x?

Answer 9

oh so I was wrong then its this one sin x cos2x - sin3x + cos3x I knew it was this one or the other one because the other choices the xs arnt correct right

Answer 10

or no it was the coeff that weren't right

Answer 11

@dpasingh

Answer 12

Ok let us solve it again.

Answer 13

sin 3x = sin (2x+x) = sin 2x cos x+ cos2x sinx
Let us know the reasons first.

Answer 14

since sin2x = 2 sinx cosx

and \[\cos2x = 1- 2\sin^2x\]
Now substituting these values, we find.

Answer 15

sin 3x = sin (2x+x) = sin 2x cos x+ cos2x sinx

\[= 2 sinx \times cosx \times \cos x+ (1-2\sin^2x) sinx\]

\[= 2 sinx \times \cos^2x + sinx-2\sin^2x \times sinx \]

\[= 2 sinx \times \cos^2x + sinx-2\sin^3x \]

\[= 2 \cos^2x sinx + sinx-2\sin^3x \]

Answer 16

I was littele bit wrong and thats why was not able to get the answer.

Answer 17

@llkramer

Answer 18

haha so I was write in the beginning xD?

Answer 19

@dpasingh

Answer 20

@llkramer

Answer 21

im here

Answer 22

so I was write in the begging?

Answer 23

i'm also here

Answer 24

so I was write

Answer 25

Yes, correct

Answer 26

?

Answer 27

sweet thank you

Answer 28

actually i used wrong formula in between

Answer 29

just submitted it and I got it right

Answer 30

Congrats