Re: Derivatives of Trig Functions

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Posted by mfrizza on October 21, 2002 at 18:38:30:

In Reply to: Re: Derivatives of Trig Functions posted by T.Gracken on October 21, 2002 at 16:25:07:

Thanks, I definitely see what to do now. I do have a quick follow-up regarding...

y' = [-2sin(x) - sin^2(x) - cos^2(x)]/[4 + 4sin(x) + sin^2(x)]

you factored out a -1 of the numerator. I'm confused as to what happened with the -2sinx. If you factor out a -1, wouldn't it be 2sinx + sin^2x + cos^2x / denom?

: : Would someone please take a look at this and tell me where I've gotten off track...?

: : Find the points on a curve y = (cos x)/(2 + sin x) at which the tangent is horizontal.

: : (using quotient formula), y' = [(2 + sin x)(-sin x) - (cos x)(cos x)] /(2 + sin x)^2

: first step looks good, but be careful when simplifying numerator...
:
: y' = [-2sin(x) - sin^2(x) - cos^2(x)]/[4 + 4sin(x) + sin^2(x)]

: so y' = -[sin(x)+1]/[2+sin(x)]2 ...leave denominator factored (my preference). I just factored out the -1 from numerator and used the pythagorean identity to rewrite the sin2(x) + cos2(x) as 1.

: now, possible points on curve at which tangent is horizontal occur when y' = 0, so the "possible" points where tangent is horizontal occur for values of x such that sin(x)+1=0 ...fraction is equal to zero when numerator is equal to zero (and denominator is not).

: this should lead to x = 3pi/2 + 2pi*n (n an integer)

: hope that helps.
:

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