Question

Evaluate :
tgx= 3/4 ; sin^3 2x- cos^3 2x / 2+4sinx

Answer 1

@Mertsj we need some help

Answer 2

what does it mean tgx=3/4

Answer 3

tangent X= 3/4

Answer 4

The abbreviation for tangent is tan

Answer 5

sorry(

Answer 6

any idea how to solve it?

Answer 7

\[\frac{\sin ^3(2x)-\cos ^3(2x)}{2+4\sin x}=\frac{(\sin (2x)-\cos (2x))(\sin ^2(2x)+\sin (2x)\cos (2x)+\cos ^2(2x)}{2+4\sin x}\]
\[\frac{(\sin (2x)-\cos (2x))(1+\sin (2x)\cos (2x))}{2+4\sin x}=\]

Answer 8

\[\frac{(2\sin xcos x-(1-2\sin ^2x))(1+(2\sin x \cos x)(1-2\sin ^2x))}{2+4\sin x}=\]

Answer 9

\[\frac{(2\sin x \cos-1+2 \sin ^2x)(1+2\sin x \cos x-4\sin ^3x \cos x)}{2+4\sin x}\]

Answer 10

If tan x = 3/4 then sin x = 3/5 and cos x = 4/5
so you can substitute and finish it.

Answer 11

okay!) Mertsj, thank you a lot!!!) that's gonna be on my exam tmrw)

Answer 12

yw

Answer 13

You might be able to make it a bit easier by saying: sin(2x)=2sinxcosx=2(3/5)(4/5)=24/25
And cos(2x)=1-2sin^2x=1-2(3/5)^2=1-2(9/25)=1-18/25=7/25

Answer 14

Then you could substitute into this step:
\[\frac{(\sin (2x)-\cos (2x))(1+\sin (2x)\cos (2x))}{2+4\sin x}\]