# Precalculus with Trigonometry

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Prove or disprove the following Identities:

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

sin raised to the 4 (theta) - cos raised to the 4 (theta) = sin squared (theta) - cos squared (theta)

cos (x+(pi)/(6)) + sin (x - (pi)/(3)) = 0

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

Sorry if you don't understand anything

• Precalculus with Trigonometry -

For the first problem:

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

A few identities for negatives:

cos(-x) = cos(x)
sin(-x) = -sin(x)

Therefore:

cos(x) - [-sin(x)] = cos(x) + sin(x)

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

Some identities:
cos(x+y) = cos(x)cos(y) - sin(x)sin(y)
cos(x-y) = cos(x)cos(y) + sin(x)sin(y)

Multiplying using both identities:

[cos^2(x) cos^2(y)] - [sin^2(x) sin^2(y)]

Next, use the identity:
sin^2(x) + cos^2(x) = 1

[1-sin^2(x)][1-sin^2(y)] - [sin^2(x) sin^2(y)]

Multiply [1-sin^2(x)][1-sin^2(y)]:

1 - sin^2(y) - sin^2(x) + [sin^2(x) sin^2(y)] - [sin^2(x) sin^2(y)]

We are left with this:
1 - sin^2(y) - sin^2(x)

Which equals this:
cos^2(x) - sin^2(y)

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I'll stop there. I hope this helps.