How to differentiate (sinx)^2?

  • Context: High School 
  • Thread starter Thread starter strokebow
  • Start date Start date
  • Tags Tags
    Differentiate
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
11 replies · 69K views
strokebow
Messages
122
Reaction score
0
How do you differentiate the likes of (sinx)^2

thanks
 
Physics news on Phys.org
Alternatively, you can recall / derive the power reduction formulae such as;

[tex]\sin^2\theta = \frac{1 - \cos 2\theta}{2}[/tex]

These are especially useful when integrating such functions.
 
Hootenanny said:
Alternatively, you can recall / derive the power reduction formulae such as;

[tex]\sin^2\theta = \frac{1 - \cos 2\theta}{2}[/tex]

These are especially useful when integrating such functions.

Differentiating, not integrating. :biggrin:
 
I Think Hootenanny was in fact intended to use that to simplify the differentiation, if I'm reading his last sentence correctly >.<...Well anyway It doesn't really help very much because we still have to use the chain rule on the cos 2theta.
 
Gib Z said:
I Think Hootenanny was in fact intended to use that to simplify the differentiation, if I'm reading his last sentence correctly >.<...Well anyway It doesn't really help very much because we still have to use the chain rule on the cos 2theta.
Not at all, I meant that the formulae are useful for differentiation, but more so for integration. It is true that to differentiate you may have to use the chain rule for both forms but I find it easier to remember that;

[tex]\frac{d}{dx}\sin(ax) dx = a\cos(ax)[/tex]

[tex]\int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C[/tex]

Rather than remembering the results for the sin2x etc. In any event applying the chain rule to something of the form sin(ax) is somewhat simpler than applying it to something of the form sin2x don't you think?
 
Last edited:
Not always, usaully one would like an answer without double angled arguments, so they would have to know the expansion of cos(2theta) which isn't as easy as bringing a power down times the derivative of sin x.
 
Gib Z said:
Not always, usaully one would like an answer without double angled arguments, so they would have to know the expansion of cos(2theta) which isn't as easy as bringing a power down times the derivative of sin x.
Fair point perhaps, but I've never come across a case where a single angle argument is preferable to a double angle. In any case, the double angle form is certainly much easier to integrate.
 
Of course there all equivalent, but I always prefer putting my answers in terms in single angled arguments. In the end it makes very little difference, maybe 5 seconds working time.
 
= 2sinx cosx
= sin2x :zzz:
 
DAKONG said:
= 2sinx cosx
= sin2x :zzz:

ditto..


w00t 1st post :P