# Compound Angles ProofWatch

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#1
Hi, ive been reading the proof for compound angle Sin(A+B) and i dont follow all of the proof.

So far ive understood:

Sin (A+B) = TR/OR
= (TS+SR)/OR
= (PQ+SR)/OR

but then it goes

=(PQ/OQ)X(OQ/OR)+(SR/QR)X(QR/OR)

which then equals SinACosB+CosASinB

Its just that last stage where you multiply them- why??!?!!?!?!

0
14 years ago
#2
i always thought that equation was derived using complex numbers.
0
#3
well this is in the P2 book and Core Maths by L Bostock and S Chandler

I mean- do we have to know how to prove it, or can we just memorise them?
0
14 years ago
#4
I don't think you need to know the proof.

(PQ+SR)/OR
=PQ/OR + SR/OR

Then you do some clever multiplying (so you don't actually change anything but you have it in a form from which you can replace each fraction with a trig identity):

=(PQ/OQ)(OQ/OR) + (SR/QR)(QR/OR)
=SinACosB+CosASinB
0
14 years ago
#5
(Original post by faa)
Hi, ive been reading the proof for compound angle Sin(A+B) and i dont follow all of the proof.

So far ive understood:

Sin (A+B) = TR/OR
= (TS+SR)/OR
= (PQ+SR)/OR

but then it goes

=(PQ/OQ)X(OQ/OR)+(SR/QR)X(QR/OR)

which then equals SinACosB+CosASinB

Its just that last stage where you multiply them- why??!?!!?!?!

It's just using the obvious fact that OQ/OQ = 1 and QR/QR = 1 and multiplying each term by one of these which doesn't change it as it's just multiplying by 1.

Mikw - there's a nice (in my opinion!) way of proving both the sine and cosine compound angle formulae by multiplying rotation matrices for angles A and B and setting the result equal to the rotation matrix for angle A+B which kills 2 birds with 1 stone.
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