Question

Rewrite with only sin x and cos x.

sin 3x

Answer 1

\[\text{TrigExpand}[\sin (3 x)]=3 \sin (x) \cos ^2(x)-\sin ^3(x) \]

Answer 2

sin(A+B) = sin Acos B+cos A sin B

sin 3x = sin (2x+ x)
= sin 2x cos x + cos 2x sin x
= sin (x+x) cosx + cos (x+x) sin x
= (sin x cos x + cos x sin x) cosx + (cos x cos x- sin x sin x) sin x
\[=\sin x \cos ^{2}x+\cos ^{2}x \sin x +\cos ^{2}x sinx - \sin ^{3}x\]

Answer 3

\[=sinx (1 - \sin ^{2}x)+(1 -\sin ^{2}x)\sin x + ( 1-\sin ^{2}x)\sin x -\sin ^{3}x\]

Answer 4

\[=\sin x - \sin ^{3}x +\sin x - \sin ^{3}x+ \sin x - \sin ^{3}x-\sin ^{3}\]

Answer 5

\[= 3\sin x -3 \sin ^{3}x\]

Answer 6

I'm a little late seeing this and I appreciate the help but none of these were the answer choices.

Answer 7

\[@alexh107\\sin3x= \sin2(\frac{ 3x }{ 2 })\ use\ identity\ \sin2A= 2sinA.cosA\\ here\ A= \frac{ 3x }{ 2 }\\ \therefore\ \sin3x= 2\sin \frac{ 3x }{ 2 }\cos \frac{ 3x }{ 2}\]

Answer 8

These are the choices I have (ignore the marked one.)

Answer 9

I decided to stick with the answer I had originally guessed which was\[2\cos^2x \sin x + \sin x - 2\sin^3x\]for anyone wondering or anyone who searches this in the future. It turned out to be correct. I looked it up and most people said there maybe something wrong with the question which might be why people were getting different answers. If anyone can explain it to me I'd appreciate it.

Answer 10

@alexh107 getting only last one marked.

Answer 11

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

Answer 12

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