Derivative of y = sin x * cos x


by TristanH
Tags: calculus, product-to-sum, trig, trig identity
TristanH
TristanH is offline
#1
Jun25-08, 09:32 PM
P: 12
I'm working through Kline's Calculus book, and am at the chapter on Integration on Differentiation of Trig Functions. A question asks to find the derivative of:

(I've labeled all equations for easy reference)

(1) y = sin x * cos x

unlike the solution guide which advocates using the product rule, I decided to use the product to sum trig identity and work from there. Unfortunately it looks like my answer is incorrect, and I'd like to know why.

So, using the trig identity:

(2) sin a * cos b = (sin(a+b) + sin(a-b)) / 2

I computed:

(3) y = (sin(x + x) + sin(x - x)) / 2

then,

(4) y = (sin(2x) + sin(0)) / 2

which simplifies to:

(5) y = (sin(2x)) / 2

Then By the chain rule:

(6) y' = (2 * cos(2x)) / 2

which yields:

(7) y' = cos(2x)

yet the solution guide has

(8) y' = -sin^2(x) + cos^2(x) which is clearly different.

Where did I go wrong?
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Defennder
Defennder is offline
#2
Jun25-08, 09:34 PM
HW Helper
P: 2,618
They are the same.

[tex]\cos (2x) = \cos^2 (x) - \sin^2(x) [/tex].
TristanH
TristanH is offline
#3
Jun25-08, 09:44 PM
P: 12
FYI - I redid the problem using the product rule and got the same answer as the text. However, why does the identity method not work?

TristanH
TristanH is offline
#4
Jun25-08, 09:46 PM
P: 12

Derivative of y = sin x * cos x


Defender - so it is! I missed that one. Thanks!
rocomath
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#5
Jun25-08, 09:59 PM
rocomath's Avatar
P: 1,757
Make it easier: [tex]\sin x\cos x=\frac{\sin{2x}}{2}[/tex]


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