# differential trig problem

• Apr 30th 2009, 11:27 PM
slaypullingcat
differential trig problem

Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
Thanks all the same.
• Apr 30th 2009, 11:40 PM
Spec
Here's a good thread about the math code this forum uses: http://www.mathhelpforum.com/math-he...-tutorial.html

$sin^4xcos^3x = sin^4x(1-sin^2x)cos x = cosx(sin^4x-sin^6x)$

The last part is easy to integrate since $\int cosxsin^{n-1}x dx = \frac{1}{n}sin^nx$
• Apr 30th 2009, 11:44 PM
redsoxfan325
Quote:

Originally Posted by slaypullingcat

Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
Thanks all the same.

Show that: $\frac{d}{dx}\left[\frac{1}{5}\sin^5(x)-\frac{1}{7}\sin^7(x)\right] = \sin^4(x)\cos^3(x)$

$\frac{d}{dx}\left[\frac{1}{5}\sin^5(x)\right] = 5\cdot\frac{1}{5}\sin^4(x)\cdot\cos(x) = \sin^4(x)\cos(x)$

$\frac{d}{dx}\left[\frac{1}{7}\sin^7(x)\right] = 7\cdot\frac{1}{7}\sin^6(x)\cdot\cos(x) = \sin^6(x)\cos(x)$

So the derivative is $\sin^4(x)\cos(x)-\sin^6(x)\cos(x) = \sin^4(x)\cos(x)\cdot(1-\sin^2(x))$ $= \sin^4(x)\cos(x)\cdot\cos^2(x) = \sin^4(x)\cos^3(x)$