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sinx cosx/tanx=sinx cosx cosx/sinx
=cosxcosx
=cos^2x
=1-sin^x

sorry...1-sin^2x

how does sinxcosx/tanx = sinx?

It doesn't. It equals cos^2x

just kidding, i understand how you wrote it now

I like how you stalked me then...lol

haha sorry. i got confused.

1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)

Can you type out the question using the equation editor?

\[1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x\]

okay

in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1

Yeah...I was thinking that.

yeahh.

Okay, it's easy now...

\[=2\cot^2 x\]

Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]

Once you're done, let me know if it works for you or not.

okay, i understand that one

\[\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2\]

how did you identify (1)?

is x. I'll draw a pic.

yeah, i get it now. i just had to go through it all

Okay

\[\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]

sin(x) is subtracted from sin(x) / sec(x) in the second equation.

Okay, sorry, had to do something. Here we go...

ok, yep.

got it so far..

Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.

ok

find exact value of (-pi/12)

function?

sin(-pi/12)

sorry

square root of 3 over 4?

Do you want to try that and see how you go. I need a shower.

You can read off your sine, cosine and tangents from them.

You could also use the combination,\[\frac{\pi}{4}-\frac{\pi}{3}\]