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November 25th, 2011, 03:17 PM   #1
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Simplying Trig

(2cosx+5sinx)^2 + (5cosx-2sinx)^2
The answer has to be in numerical value

This is what I have so far
=2cos^2x + 2(2cosx5sinx) + 5sin^2x + 5cos^2x -2(5cosx2sinx) + 2sin^2x
=2cos^2x + 4cosx10sinx + 5sin^2x + 5cos^2x -10cosx4sinx + 2sin^2x
=7cos^2x + 7sin^2x + 4cosx10sinx - 10cosx4sinx

Not sure what to do from here.
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November 25th, 2011, 03:25 PM   #2
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Re: Simplying Trig

(2cos(x) + 5sin(x))^2 + (5cos(x) - 2sin(x))^2 = 4cos²(x) + 20cos(x)sin(x) + 25sin²(x) + 25cos²(x) - 20cos(x)sin(x) + 4sin²(x)

Simplify and rearrange:

= 4cos²(x) + 4sin²(x) + 25cos²(x) + 25sin²(x) = 4(cos²(x) + sin²(x)) + 25(cos²(x) + sin²(x))

apply Pythagorean identity, cos²(x) + sin²(x) = 1,

= 4(1) + 25(1) = 29
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November 25th, 2011, 06:02 PM   #3
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Re: Simplying Trig

Another way to proceed would be to use a linear combination identity:







To simplify the second factor, let :



We know:

thus:



giving us:



Since we may now write:

and going back to the original problem, we now have:



In this problem, a = 2 and b = 5, thus it is evaluates to:



The method used by greg1313 is by far more straightforward, I merely wanted to demonstrate an alternative...
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November 25th, 2011, 09:43 PM   #4
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There were two mistakes in the original work.

The first was failing to square the coefficients 2 and 5 to get 4 and 25.

The second was changing 2(2cosx5sinx) to 4cosx10sinx. Multiplication can be distributed over addition, but not over multiplication. The correct value would be 4cosx5sinx or 20cosxsinx.

With these corrected, 29cos²x + 29sin²x could have been obtained, which is 29 (as cos²x + sin²x = 1).
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