Question

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 1

rewrite tan and cot in terms of sin cos

Answer 2

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2x

Answer 3

(sinx)(sinx/cosx cosx- cosx/sinx cosx)

Answer 4

what do i do next

Answer 5

\[\large\rm \sin(x)(\frac{\sin x}{cosx} * cosx -\frac{\cos x }{sinx } *\cos x) \]
first solve the parentheses
sinx/cos times cos x = ?

Answer 6

sinx

Answer 7

\[sinx(sinx-\cos^2x /sinx)\]

Answer 8

right cos cancel out \[\large\rm \sin(x)(\frac{\sin x}{\cancel{cosx}} *\cancel{ cosx} -\frac{\cos x }{sinx } *\cos x) \]
\[\large\rm \sin(x)( sin x -\frac{\cos x }{sin x } *\cos x) \]
now you can distribute
by sinx

Answer 9

right

Answer 10

\[\sin^2x-\cos^2x\]

Answer 11

\[\rm sinx ( sinx - \frac{\cos^2x}{sinx})\]
multply both terms in the parentheses by sin x

Answer 12

oh wow nice that's right

Answer 13

what do i do from there

Answer 14

now we can use the identity \[\huge\rm sin^2 \theta + \cos^2 \theta = 1\] solve this identity for sin^2x

Answer 15

\[\sin^2x=1-\cos^2x\]

Answer 16

i get it now thank you

Answer 17

nice !!
we can replace sin^2x with 1-cos^2x
\[\huge\rm \color{Red}{sin^2x}-\cos^2x\]
\[\huge\rm \color{Red}{(1- \cos^2x)}-\cos^2x\]
now combine like terms

Answer 18

nice good work ! keep it up! :=))