View Full Version : Derivatives of trigonometric functions
thinay
Aug 11, 2009, 04:22 AM
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
answer: sin^2 x
2.) 2tan 1/2 x - x
answer: tan^2 1/2 x
galactus
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
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
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
Aug 11, 2009, 07:52 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.
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)