thinay Posts: 40, Reputation: 2 Junior Member #1 Aug 11, 2009, 04:22 AM
derivatives of trigonometric functions
please help me with this... :) Thanks in advance! When I tried to answer this last two numbers, the answer in the book with mine. And I think my answers are wrong. Here it is:

1.) y = 1/2 x - 1/4 sin2x
2.) 2tan 1/2 x - x
answer: tan^2 1/2 x
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Aug 11, 2009, 04:30 AM
1.) $y = \
frac{x}{2} - \frac{sin(2x)}{4}$

answer: $sin^{2}(x)$
The derivative of $\frac{x}{2}$ is just 1/2.

The derivative of sin(2x) is 2cos(2x)

So, we have $\frac{1}{2}-\frac{cos(2x)}{2}=\frac{1-cos(2x)}{2}$

Which is an identity. $\frac{1-cos(2x)}{2}=sin^{2}(x)$
 thinay Posts: 40, Reputation: 2 Junior Member #3 Aug 11, 2009, 04:43 AM
Oh.. I see.. my answer is correct which is 1-cos2x/2.. I just need to simplify it to get sin^2x..

In #2, the answer that I got is sec^2 1/2 x.. Its wrong.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #4 Aug 11, 2009, 04:52 AM
You're close about #2. You got sec, which is correct. Another identity.

$sec^{2}(x)-1=tan^{2}(x)$

Don't forget that the derivative of x is 1.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 Aug 11, 2009, 07:52 AM
Originally Posted by thinay
Oh.. i see.. my answer is correct which is 1-cos2x/2.. i just need to simplify it to get sin^2x..

In #2, the answer that i got is sec^2 1/2 x.. its wrong.
Ok, I don't know either that identity, but I do know that:

$cos(2x) = 1-2sin^2(x)$

So, just rearrange, you'll have:

$cos(2x)-1 = -2sin^2(x)$

$1-cos(2x) = 2sin^2(x)$

$\frac{1-cos(2x)}{2} = sin^2(x)$

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