Precalculus with Trigonometry
posted by Natalie .
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(xy) = cos squared (x)  sin squared (y)
Sorry if you don't understand anything

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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For your last problem:
cos(x+y)cos(xy) = cos^2(x)  sin^2(y)
Some identities:
cos(x+y) = cos(x)cos(y)  sin(x)sin(y)
cos(xy) = 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
[1sin^2(x)][1sin^2(y)]  [sin^2(x) sin^2(y)]
Multiply [1sin^2(x)][1sin^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.