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C3 Trig

FullSizeRender-4.jpgCant seem to figure out the proof in the start of the question. Mark scheme sheds no light on the matter either.
Original post by Sourestdeeds
FullSizeRender-4.jpgCant seem to figure out the proof in the start of the question. Mark scheme sheds no light on the matter either.


Post what you've tried -
Original post by Sourestdeeds
FullSizeRender-4.jpgCant seem to figure out the proof in the start of the question. Mark scheme sheds no light on the matter either.


Are you sure this is C3? It seems to require C4 identities (unless non-WJEC boards include those in C3).
Original post by Hydeman
Are you sure this is C3? It seems to require C4 identities (unless non-WJEC boards include those in C3).


Its from the OCR C3 Jan 2013 paper
FullSizeRender-5.jpg

I doubt my attempt is much use! Its just I've not come across a double angle formula which is squared, so not sure how to approach.
To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution
Original post by Sourestdeeds
To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution


Remember that

cos2(a+b)=(cos(a+b))2=(cos(a)cos(b)sin(a)sin(b))2\cos^2(a+b) = (\cos(a+b))^2 = (\cos(a)\cos(b) - \sin(a)\sin(b))^2

so you've missed a term in the second line.
Reply 7
ill have a go
cos2(x45)12(cos2asin2a)=sin2x[br][br]cos2(x45)12(cos2asin2a)=1cos2x[br][br]cos2(x45)12((sin2xcos2x)sin2a)=1cos2x[br][br]cos2(x45)12(((1cos2x)cos2x)sin2a)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]2cos2(x45)1+2cos2x+sin2x=22cos2x[br][br]2cos2(x45)3+4cos2x+sin2x=0[br][br]2cos2(x45)3+4cos2x+2(cosx)(sinx)=0[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x[br][br] cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x[br][br]2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 [br][br]2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0[br][br]

Continue from here
(edited 8 years ago)
Original post by SCAR H
ill have a go
cos2(x45)12(cos2asin2a)=sin2x[br][br]cos2(x45)12(cos2asin2a)=1cos2x[br][br]cos2(x45)12((sin2xcos2x)sin2a)=1cos2x[br][br]cos2(x45)12(((1cos2x)cos2x)sin2a)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]2cos2(x45)1+2cos2x+sin2x=22cos2x[br][br]2cos2(x45)3+4cos2x+sin2x=0[br][br]2cos2(x45)3+4cos2x+2(cosx)(sinx)=0[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x[br][br] cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x[br][br]2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 [br][br]2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0[br][br]

Continue from here


Its an identity? Surely you start with the left hand side to prove the right hand side?
Reply 9
Original post by Sourestdeeds
Its an identity? Surely you start with the left hand side to prove the right hand side?


Hmm didnt see that

Works the same way though
Original post by SCAR H
ill have a go
cos2(x45)12(cos2asin2a)=sin2x[br][br]cos2(x45)12(cos2asin2a)=1cos2x[br][br]cos2(x45)12((sin2xcos2x)sin2a)=1cos2x[br][br]cos2(x45)12(((1cos2x)cos2x)sin2a)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]cos2(x45)12(12cos2xsin2x)=1cos2x[br][br]2cos2(x45)1+2cos2x+sin2x=22cos2x[br][br]2cos2(x45)3+4cos2x+sin2x=0[br][br]2cos2(x45)3+4cos2x+2(cosx)(sinx)=0[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x[br][br]cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x[br][br] cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x [br][br]2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x[br][br]2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 [br][br]2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0[br][br]

Continue from here


Original post by Sourestdeeds
Its an identity? Surely you start with the left hand side to prove the right hand side?


Yes, the problem with that kind of thing is that you're already assuming that the identity holds, so it's not a proof.

You have to start with one side and show that it's equivalent to the other.
Reply 11
Original post by Indeterminate
Yes, the problem with that kind of thing is that you're already assuming that the identity holds, so it's not a proof.

You have to start with one side and show that it's equivalent to the other.


Yes
Did it! :smile:

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Im pretty ill at the moment, making some hilarious mistakes haha. Apparently the LHS is the same as the LHS. Perfect logic

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