Question

simplify 2sinx^7x/cos^3x *[cosx/2sinx]^3 please help

Answer 1

\[(\frac{ 2\sin^7x }{ \cos^3x }) (\frac{ cosx }{ 2sinx })^3 \]

like this?

Answer 2

yes

Answer 3

first simplify the cube
\[(\frac{ 2\sin^7x }{ \cos^3x })(\frac{ \cos^3x }{ 2\sin^3x }) \]

and do you know how to multiply fractions?
i give you the general formula for it
\[(\frac{ a }{ b })(\frac{ c }{ d }) = \frac{ ac }{ bd } \]

multiply that for me

Answer 4

2sin^7x*cos^3x/cos^3x*2sin^3x

Answer 5

ok good
now notice that
\[\frac{\ 2cos^3x }{\ 2cos^3x } = 1 \]
so you can cancel that out and you are left with

\[\frac{ \sin^7x }{ \sin^3x }\]

Answer 6

now notice that \[\sin^7x = (\sin^4x)(\sin^3x)\]
so you can cancel out the denominator

so whatever you are left with is the simplified form of the original equation

Answer 7

so this is the final answer = 1+ sin^4xsin^3x

Answer 8

uhhh its just \[\sin^4x \]

Answer 9

oky
i got it