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\[\left( \cos2x \over \sin x \right) = \left( \cot^2x-1 \over \csc x \right)\]
i got it until (cos^2 over sinx) - (sinx) after your second step i multiplied deno and numerator by sinx. but how did you get cos^2-sin^2x over sinx and how is it same as cos^2x over sinx?
Because cos(2x) = cos^2(x)-sin^2(x)
Look at your trig double angle formulas
ohh aha! thanks. where did sin^2x come from
yea i meant shouln't you get cos^2-sinx over sinx instead of cos^2x -sin^2x over sinx
i am so sorry.. shouldn't you be dividing them by sinx not sin^2x? hahah :P
thanks! i always thought |dw:1325731485027:dw|
hehe i learned something 8p im always so confused .. : Sdfksj skJ anyway okay keep going please
ohhhhhhh!!!!!!!!!!!!!!! you can always work out right side instead of the left side too right? thanks!!!!!!!!!!!!!!! :P
Thanks alot for being patient with me :) and your explanation is great and quite fast! :)
You can work on either side when proving an identity or both sides. You just can't work across the equal sign like adding the same thing to both sides or something like that.
Well I could go faster if I knew how to make fractions with the equation thing.
you just type out 'over'
uw. p.s. are you a tutor? like to help students? anywho when you see identity questions how do you know what to do and which side to solve first?
I used to teach math in high school for 32 years and now I am retired and tutor students who come to my house. I can usually solve identities.
oh wow, i see. well thanks again :) your help is always appreciated! :)
i got stuck again.. \[\left( cosx \over 1+sinx \right) = \left( 1-sinx \over cosx \right)\] what should be the nxt step ?