SOLUTION: Please help me simplify the following: 4sin2x*cos2x 1-sec^2*x cos^2*x-sin^2*x Thank you

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Question 205235: Please help me simplify the following:
4sin2x*cos2x
1-sec^2*x
cos^2*x-sin^2*x
Thank you

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
4sin(2x)*cos(2x)
It takes time and practice to start "seeing" how expressions can fit the patterns of the identities. For this first one, I "see" a sin(2A) pattern. There is an identity: sin(2A) = 2sin(A)cos(A). The most important part of this pattern is that you have a product of sin and cos of the same thing. I use a vague word like "thing" on purpose. It does not matter what the arguments to sin and cos are as long as they are the same. The "2" in front of it all can always be "maufactured" with some creative factoring.
In order to see how 4sin2x*cos2x fits the pattern, first I will factor out a 2:
2(2sin2x*cos2x)
You might now see the pattern. If not, then let A be the argument to sin and cos, which is "2x". Substituting A in we get
2(2sin(A)*cos(A))
Now the expression in the parentheses matches the pattern exactly. We can now substitute in sin(2A) for the expression:
2(sin(2A))
And finally we need to replace the A with 2x giving:
2(sin(2(2x)) = 2sin(4x)

Here's an example of how you can "manufacture" the proper coefficient. Let's say we are trying to simplify 5sin(x)cox(x). In order to patch the sin(2A) pattern, I need a 2 in fron. To get the 2 in front of 5sin(x)cos(x), factor out 5/2!!:
5sin%28x%29cos%28x%29+=+%285%2F2%29%282sin%28x%29cos%28x%29%29
Now I can us the pattern and get
5sin%28x%29cos%28x%29+=+%285%2F2%29%282sin%28x%29cos%28x%29%29+=+%285%2F2%29sin%282x%29
1-sec^2*x
I assume you mean: 1-sec^2(x) (There is no multiplication involving x here (not matter how the Algebra.com software may make it look like a multiplication). sec%5E2%28x%29 means sec%28x%29+%2A+sec%28x%29
Another identitiy has sec^x and 1 in it:
tan^2(x) + 1 = sec^2(x)
If we substitute in for sec^(x) we get:
1 - (tan^2(x) + 1)
and simplifying we get:
1 - tan^2(x) + 1
-tan^2(x)

cos^2*x-sin^2*x
Again I assume you mean: cos^2(x) - sin^2(x)
This is exactly one of the identities:
cos(2x) = cos^2(x) - sin^2(x)
so your simplified expression is:
cos(2x)