Help - Trig Question from Year 12 Cambridge 3U Further Trig (1 Viewer)

[lachy95]

Hey,

Can anyone help me out on this question?

Its 13k in Exercise 2A of the Year 12 3U Cambridge Book. The question is as follows

cos²A.cos²B-sin²A.sin²B=(cos2A+cos2B)/2

Can't get it!

Cheers

 

[Sanjeet]

What's the question? Prove?

 

[kkmok]

cos^2 a * cos^2 b - sin^2 a * sin^2 b
= (cos a cos b - sin a sin b)(cos a cos b + sin a sin b)
= cos(a+b)*cos(a-b)

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cos (x+y) = cos x cos y - sin x sin y
cos (x-y) = cos x cos y + sin x sin y

therefore cos (x+y) + cos(x-y) = 2 cos x cos y

let x = a+b, y = a-b
cos (a+b) * cos(a-b)
= 1/2 * [cos(a+b+a-b)+ cos(a+b-a+b)]
= 1/2 [cos 2a + cos 2b]

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obviously using the identity cos x cos y = 1/2 [cos(x+y) + cos(x-y)] straight is out of the 3unit maths syllabus ( it's in the 4unit syllabus)
I am not bother to do the ext1 maths way but this is probably valid in a 3unit style exam

 

[Sanjeet]

^ I don't think that identity is in the syllabus anymore.

 

[lachy95]

yeah, it was prove, Sanjeet. thanks heaps kkmok. But Sanjeet, you're right, never heard of that identity cos x cos y = 1/2 [cos(x+y) + cos(x-y)]

 

[iBibah]

Hint:

LHS= cos(2A+2B)

Now prove that = RHS. ( remember you can work from both sides not just LHS)