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Ber [7]
3 years ago
6

NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math.Don't guess please.

Mathematics
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

the answer is A........

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Please show that;<br><br> Sin(x+y)cos(x-y) = sin x cos x + sin y cos y
bulgar [2K]

Apply the rules of sum/difference for trigonometric functions:

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

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

Multiply the two expressions:

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

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

\sin^2(x) \cos(y)\sin(y) + \sin(x) \cos^2(y)\cos(x)+ \cos(x) \sin^2(y)\sin(x) + \cos^2(x) \sin(y)\cos(y)

You can factor \cos(y)\sin(y) from the first and last term, and \sin(x)\cos(x) from the two middle terms: you have

\cos(y)\sin(y)(\sin^2(x)+\cos^2(x)) + \sin(x) \cos(x)(\cos^2(y)+\sin^2(y))

Since \sin^2(x)+\cos^2(x)=1, the identity is proven.

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Number 9 <br> Math problem
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This needs more information
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Someone pls help me ill give out brainliest pls don’t answer if you don’t know
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Can it be decimal?

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