how do you prove this is correct cos(x-y) = cosx cosy + sinx siny?

Answer 1

cos(x-y) = cos(x+(-y))
= cosx cos(-y) - sinx sin(-y)
= cosx cosy - sinx -sin(y)
= cosx cosy + sinx siny

Answer 2

Cosx Cosy

Answer 3

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RE:
how do you prove this is correct cos(x-y) = cosx cosy + sinx siny?

Answer 4

Using the identity for cos(x+y) without first deriving it is I think is not really a valid proof. Because the question is to PROVE the identity and not DERIVE it, a simpler approach that does not involve deriving cos(x+y) is to work from the right side of the equation. Using the exponential forms:

cos(x) = (e^ix + e^-ix)/2 and sin(x) = (e^ix - e^-ix)/(2i)

expand cos(x)cos(y) + sin(x)sin(y). Multiply out the terms, and you will get to the result

cos(x)cos(y) + sin(x)sin(y) = [e^i(x-y) + e^-(x-y)]/2

or

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

Answer 5

Sinxsiny

Answer 6

Plug in some numbers for x and y then run it through the calculator.
Like X=1 Y=1

Answer 7

There are proofs for the sin(x+y) and cos(x+y) versions here
http://en.wikipedia.org/wiki/Trigonometric_identity#sin.28x_.2B_y.29_.3D_sin.28x.29_cos.28y.29_.2B_cos.28x.29_sin.28y.29
And then the next section explains how to use those two to prove cos(x-y).

Answer 8

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

We start with the identity

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

Which is proved geometrically. See http://en.wikipedia.org/wiki/Trigonometric_identity

Substitute y with −y into the cos(x + y) formula:

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

Sine is an odd function and cosine is an even function, so

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

Answer 9

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The eight trig identities that you should know and know how to use before your exam are: 1) sin(x+y) = sin(x)*cos(y) + sin(y)*cos(x) 1a) sin(x-y) = sin(x)*cos(y) – cos(x)* sin(y) 2) sin(2x) = 2*sin(x)*cos(x) 3) cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) 3a) cos(x-y) = cos(x)*cos(y) + sin(x)*sin(y) 4) cos(2y) = cos^2(y) - sin^2(y) 5) cos(x) = sqrt(1 - sin^2(x)) 6) sin(y) = sqrt(1 - cos^2(y)) I am stating this based on my own experience teaching trig courses. --------------------- The following three of those eight trig identities are relevant to this problem: 1) sin(x+y) = sin(x)*cos(y) + sin(y)*cos(x) 1a) sin(x-y) = sin(x)*cos(y) - sin(y)*cos(x) 2) sin(2x) = 2*sin(x)*cos(x) --------------------- So now the problem becomes: sin(x)*cos(y) + sin(y)*cos(x) - sin(x)cos(y) + sin(y)*cos(x) ?= 2sin(y)*cos(x) and after canceling and adding together the left hand side of this equation you get: 2*sin(y)*cos(x) ?= 2*sin(y)*cos(x) and they are equal. End .

Answer 10

Using Euler's formula(e^(i*x)=cos(x)+sin(x)i) it can be shown that cos(z)=[e^(i *z)+e^(-i*z)]/2

Using this
we can prove that cos(x+y)=cos(x)*cos(y)-sin(x)*sin(y)
1) cos(x+y)= [e^( i * (x+y)) + e^(-i * (x+y))]/2
2) 2*cos(x+y)= e^( i * (x+y)) + e^(-i * (x+y))
3) =e^(i x)*e^(i y)+e^(-i x)* e^(-i y)
using Euler's formula we expand each power
4) =(cos(x)+sin(x)i)(cos(y)+sin(y)i)+
(cos(-x)+sin(-x)i)(cos(-y)+sin(-y)i)
distribute and rearrange the terms
5) =cos(x)cos(y) - sin(x)sin(y)+cos(-x)cos(-y) - sin(-x)sin(-y)+cos(x)sin(y)i+cos(y)sin(x)i+ cos(-x)sin(-y)i+cos(-y)sin(-x)i
using the idenities sin(-x)=-sin(x) and cos(-x)=cos(x) the expression becomes
6) =cos(x)cos(y) - sin(x)sin(y)+cos(x)cos(y) - sin(x)sin(y)+cos(x)sin(y)i+cos(y)sin(x)i-cos(x)sin(y)i-cos(y)sin(x)i
7) =2[cos(x)cos(y)-sin(x)sin(y)]
divide both sides by 2
8)2 cos(x+y)/2 =2[cos(x)cos(y)-sin(x)sin(y)]/2 ->cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
replacing y with -y yields the formula
9) cos(x+(-y))=cos(x)cos(-y)-sin(x)sin(-y)
idenities sin(-x)=-sin(x) and cos(-x)=cos(x) again we show that
10)cos(x+(-y))=cos(x)cos(y)+sin(x)sin(y)