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C3 trig double angle and binomial expansion

I had a question in the c3 book, i had to write the following in its simplest form and with only one trig function, Here it is

sin4θ2sin2θcos2θ+cos4θsin^4\theta-2sin^2 \theta cos^2 \theta+cos^4 \theta after finding the answer cos22θ cos^2 2 \theta (just barely, i actualy very much doubted this answer because of the way the original function looked i.e the 2sin2θcos2θ-2sin^2\theta cos^2 \theta but it turned out to be the right answer, doesn't this look like binomial exapnsion? the onl thing stoping it from looking exacly like binomial is the - being there instead of + +. another reason i did not think my answer was right was because cos2θ=cos2θsinθ cos 2 \theta =cos^2 \theta-sin^ \theta not the + + that i had in the original function.
(edited 13 years ago)
Reply 1
Your answer's right. Like, what do you get when you square (uv)2(u-v)^2? So set u=cosθ,v=sinθu=\cos \theta, v=\sin \theta and see what happens.
Reply 2
cos2θ2sinθcosθ+sin2θ cos ^2 \theta - 2sin\theta cos\theta +sin^2 \theta thats one to remember, only thing is ive forgotten how i go the answer i think i tried to sqaure cos2θsinθ cos^2 \theta -sin^ \theta to get it into the form of the original function but i kept skipping the working as i thought it was going to be wrong, so i made assumption about it.
(edited 13 years ago)
Reply 3
Original post by Core
cos2θ2sinθcosθ+sin2θ cos ^2 \theta - 2sin\theta cos\theta +sin^2 \theta


Sorry, I meant u=cos2θu=\cos^2 \theta and v=sin2θv=\sin^2 \theta; my bad.
Reply 4
Original post by nuodai
Sorry, I meant u=cos2θu=\cos^2 \theta and v=sin2θv=\sin^2 \theta; my bad.


No worries you still helped, It seems as if there are alot of things im ging to have to "look out for" if i want to be good at solving this problems, such a as the one in this question, fortunately with practice and help ive been getting better.
Reply 5
Original post by Core
No worries you still helped, It seems as if there are alot of things im ging to have to "look out for" if i want to be good at solving this problems, such a as the one in this question, fortunately with practice and help ive been getting better.


I think with this question it's just a case of noticing that you can factorise it. Usually the actual exam questions on trig identities are more obvious than this one, but it doesn't hurt to practice. There aren't many new tricks they can throw at you to be honest, so just practice questions like this and you'll have no problems.
Reply 6
can somene help mee Given that:

cos(x+30) = 3 cos(x-30) , prove that,

tan x = - ''root 3''/2

then prove that: (1 - cos '2theta') / sin '2theta' = tan 'theta'

and verify that 'theta' = 180 is a solution of the equation sin 2'theta' = 2 - 2 cos2'theta'

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