How do I verify the following? #cos(x+y)cos(x-y)=cos^2x-sin^2y#

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vince Share
Mar 10, 2015

Hello,

Use usual formulas :

#cos(x+y) = cos(x)cos(y) - sin(x) sin(y) = A+B#,
#cos(x-y) = cos(x)cos(y) + sin(x) sin(y) = A-B#.

So, #cos(x+y)cos(x-y) = (A+B)(A-B) = A^2 - B^2#,

it means

#cos(x+y)cos(x-y) = cos^2(x)cos^2(y) - sin^2(x)sin^2(y).#

But you know that #cos^2(T) + sin^2(T) = 1#, so,

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x)(1-sin^2(y)) - (1-cos^2(x))sin^2(y)#.

Develop and find

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x) -cos^2(x) sin^2(y) - sin^2(y) + cos^2(x)sin^2(y)#.

Simplify :

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x) - sin^2(y)#.

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Write your answer here...
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Then teach the underlying concepts
Don't copy without citing sources
preview
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Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

18
vince Share
Mar 10, 2015

Hello,

Use usual formulas :

#cos(x+y) = cos(x)cos(y) - sin(x) sin(y) = A+B#,
#cos(x-y) = cos(x)cos(y) + sin(x) sin(y) = A-B#.

So, #cos(x+y)cos(x-y) = (A+B)(A-B) = A^2 - B^2#,

it means

#cos(x+y)cos(x-y) = cos^2(x)cos^2(y) - sin^2(x)sin^2(y).#

But you know that #cos^2(T) + sin^2(T) = 1#, so,

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x)(1-sin^2(y)) - (1-cos^2(x))sin^2(y)#.

Develop and find

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x) -cos^2(x) sin^2(y) - sin^2(y) + cos^2(x)sin^2(y)#.

Simplify :

#cos^2(x)cos^2(y) - sin^2(x)sin^2(y) =cos^2(x) - sin^2(y)#.

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