Expert: Josh - 3/1/2004

Question
Dear Sir,

I am trying to solve a problem from Trigonometry but I am unable to solve it. The problem is from my Higher Maths textbook.

Prove that -

cos[(A-B)/2] - sinA.sin[(A+B)/2] =
cosA.cos[(A+B)/2]

If you can please answer this question, I will remain grateful to you for ever

Thanks in Advance.

Answer
Hello Aditya Kumar Singh,

Thanks for stating and typesetting the question so clearly. This question is not too difficult, we just need to take care with the algebra.

It is more convenient to show that
cos[(A-B)/2] = cosA.cos[(A+B)/2]+ sinA.sin[(A+B)/2] ...[#]

We will be using the following two identities:
cos(x+y)=cos(x)cos(y)-sin(x)sin(y). [ID1]
sin(x+y)=sin(x)cos(y)+cos(x)sin(y). [ID2]

Such is the art of mathematical proofs, the actual proof may omit the following section to make the argument more succinct. But during the actual construction of the proof, we make the following observation. After all, we need to know what it is, that we are aiming to show, right?
==================================================
OK. Putting x=A/2, y=-B/2 in [ID1], we see that

LHS of [#] = cos[(A-B)/2]
=cos(A/2)cos(-B/2)-sin(A/2)sin(-B/2).
=cos(A/2)cos(B/2)-sin(A/2)*[-sin(B/2)]
=cos(A/2)cos(B/2)+sin(A/2)sin(B/2)

So, here is the thought. We want to manipulate the RHS of [#] into this.
==================================================

Actual proof can start here.

Expanding the RHS, the first part,
cos(A).cos[(A+B)/2]
=cos(A/2+A/2).cos[(A+B)/2]
=[cos(A/2)cos(A/2)-sin(A/2)sin(A/2)].[cos(A/2)cos(B/2)-sin(A/2)sin(B/2)]
=[2*(cos(A/2))^2 -1] .[cos(A/2)cos(B/2)-sin(A/2)sin(B/2)]
= 2[cos(A/2)]^2 *cos(A/2)cos(B/2) -2*[cos(A/2)]^2 *sin(A/2)sin(B/2) + sin(A/2)cos(B/2) - cos(A/2)cos(B/2).
...let this expression be [#1]

the second part,
sinA.sin[(A+B)/2]
=sin(A/2+A/2).sin[(A+B)/2]
=2[sin(A/2)cos(A/2)].[sin(A/2)cos(B/2)+cos(A/2)sin(B/2)]
=2*[sin(A/2)]^2 * cos(A/2)cos(B/2) + 2*[cos(A/2)]^2*sin(A/2)sin(B/2)
...let this expression be [#2]

RHS of [#] = [#1]+[#2].....after some cancellations.....
= 2[cos(A/2)]^2 *cos(A/2)cos(B/2) +2*[sin(A/2)]^2 * cos(A/2)cos(B/2) + sin(A/2)cos(B/2) - cos(A/2)cos(B/2)
.....next, since 2[cos(A/2)]^2 + 2[sin(A/2)]^2 = 2(1)
= 2 cos(A/2)cos(B/2)+ sin(A/2)cos(B/2) -cos(A/2)cos(B/2)
= sin(A/2)cos(B/2) + cos(A/2)cos(B/2)
= cos[(A+B)/2]
=LHS of [#].

Cheers.