Question

Verify the equation:
2. (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2 x

Answer 1

1) distribute sin x
2) write tan x = sin x/ cos x, cot x = cos x/sin x.

Answer 2

Can we start with the right side?

Answer 3

Or no? Because I think I'm supposed to match right to left

Answer 4

is it asked to march right to left ? that would be not as easy as to go from left to right...

Answer 5

Yeah I know! haha but that's what it says lol

Answer 6

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

Answer 7

ok, then you can first use the formula,
1= \(\sin^2x + \cos^2x\)

Answer 8

okay so sin^2 x+cos^2 x- 2 cos^2 x

Answer 9

yes, which is sin^2x-cos^2x, isn't it ?

Answer 10

now whats done is :
\(\dfrac{\sin^2x \cos x}{\cos x} - \dfrac{\cos^2x \sin x}{\sin x} \)

Answer 11

then 1)factor out sin x from numerator of both terms,
2) write sin x/cos x = tan x in 1st term and cos x/sin x =cot x in 2nd term.
and you're done...

Answer 12

okay so would I use the tanx=sinx/cosx and cotx=cosx/sinx formulas?

Answer 13

oh you just wrote that haha

Answer 14

okay let me try

Answer 15

sure, take your time :)

Answer 16

Would it be tan^2x cos x - cot^2 x sin x? I feel like that might be wrong

Answer 17

tillwhich step are u sure, you got correct ?

Answer 18

I'm not sure how to factor out sin x

Answer 19

\(\dfrac{\sin^2x \cos x}{\cos x} - \dfrac{\cos^2x \sin x}{\sin x} \\ = \sin x [\dfrac{\sin x \cos x}{\cos x} - \dfrac{\cos^2x }{\sin x}] \\ =\sin x[\dfrac{\sin x }{\cos x} \times \cos x - \dfrac{\cos x }{\sin x} \times \cos x]\)
now ?

Answer 20

Oh! soo... sinx(tanx cosx- cot x cos x) You just made this whole verifying thing so much! Easier! I think that's where I'm messing up is factoring out! Thank you so much! :)

Answer 21

welcome and keep practising ^_^