Question

Proof:

cos² (a + b) + cos² (a - b) = 1 + cos 2a * cos 2b

Answer 1

Resolve LHS using double angle formula of cosine and then use sum of angle formula of cosine to simplify.

Answer 2

As english isn't my native language, with double angle formula you mean

LHS = cos²(2a) = 1 + cos2a? But then there's no "b" left in the equation, so I'm probably going wrong somewhere at working out the formula :/

Answer 3

see http://mathworld.wolfram.com/Double-AngleFormulas.html
convert (cos(a))^2 into cos(2a).

Answer 4

I'm sorry , i don't really get what you mean. This is all really new material to me and let's say I can't really rely on my teacher to explain it properly..

So basically the left side: cos²(a+b) + cos²(a-b) can be resolved into

= cos² a
= (1 + cos 2a)/2

right side:

worked out by the double angle formulas? That does leave me with a "b" that is not in my left side of the equation...

Answer 5

see the trouble is that on LHS the form is {cos(sum)}^2 + {cos(difference)}^2 while on RHS it is something to do with cos(double a) and cos(double b).So you need to make both sides in same form to compare.so convert cos(x)^2 terms of LHS into cos(2x) form.Here x will be a sum or difference.so thereafter use cos(m+n)=... formula.

Answer 6

that gives me: 1 + cos[2(a+b)] + 1 + cos[2(a-b)].
= 1 + cos 2a * cos 2b - sin 2a * sin 2b + 1 + cos2a * cos2b + sin2b * sin2a
= 1 + cos2a * cos2b + 1 + cos2a * cos2b
= 2 + cos2a * cos 2b + cos2a * cos2b
= 2 + 2(cos2a*cos2b)
= 1 + cos2a * cos2b

Is this correct?

Answer 7

Actually, I think I forgot the division by 2 in the formula, adding that gives me the reason to divide by 2 at the 2 + 2(cos2a * cos2b) stage, I suppose?

Answer 8

yep there should be a 1/2 in the first step itself.rest is correct..

Answer 9

thanks