RightSaidJames
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#21
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#21
(Original post by milan5baros)
thats just stating a formula.....
No it's not, it's telling you flat out that, in general terms, one does not equal the other, so you can't assume that.
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chr15chr15
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#22
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#22
(Original post by RightSaidJames)
noooooooooo lol... but they might ask you a question where someone says "Rachel says:
cos(this) + cos(that) is the same as cos(this + that), therefore cos(A) + cos(B) must always equal cos(A + B).
Prove rachel is wrong in her assumption"

All you do is choose two numbers where that doesn't work.
o i was gonna say!
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milan5baros
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#23
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#23
well I didn't understand you. But I've cleared it up now.
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grace_
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#24
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#24
Cos 165 = Cos (45 + 120)
Cos (45 + 120) = Cos 45 Cos 120 - Sin 45 Sin 120.

(If you think about it, Cos 90 = 0.
 Cos 45 + Cos 45 = \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} = \frac{2}{\sqrt 2}
So if Cos 45 + Cos 45 = Cos 90
then you're saying that  \frac{2}{\sqrt 2} = 0 which isn't true. )

Whenever you're finding Cos (one number + another number)
or Sin (one number + another number)
then you have to use the formulae for Cos(A+B) etc.
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Maths Buster
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#25
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#25
(Original post by RightSaidJames)
Because cos(A \pm B) \neq cosA \pm cosB
Just to put a spanner in the works, if
A= 180 degrees, B = 60 degrees, so that A + B = 240 degrees, then
cos[A + B] = cosA + cosB.
In general it is not true that cos[A + B] = cosA + cosB, but there are occasions when it is true. Similarly for sin[A + B] etc.
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