X = sin(a)/cos(b), find da/dx

  • Thread starter Gekko
  • Start date
  • #1
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Homework Statement



x=sin(a) / cos(b)

a+b < pi/2
a>0, b>0

show that

da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b)


The Attempt at a Solution



dx/da = cos(a) / cos(b) therefore
da/dx = cos(b) / cos(a)

=cos^3(b)cos(a) / cos^2(b)cos^2(a)

However the denominator of the desired format = cos(a+b)cos(a-b) = cos^2(a)cos^2(b)-sin^2(a)sin^2(b)

Not sure how to get rid of the sin^2(a)sin^2(b) term...
 

Answers and Replies

  • #2
34,183
5,801

Homework Statement



x=sin(a) / cos(b)

a+b < pi/2
a>0, b>0

show that

da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b)


The Attempt at a Solution



dx/da = cos(a) / cos(b) therefore
da/dx = cos(b) / cos(a)

=cos^3(b)cos(a) / cos^2(b)cos^2(a)

However the denominator of the desired format = cos(a+b)cos(a-b) = cos^2(a)cos^2(b)-sin^2(a)sin^2(b)

Not sure how to get rid of the sin^2(a)sin^2(b) term...
You are trying to show that cos(a+b)cos(a-b) = cos^2(b)cos^2(a). I am not able to find where you went wrong, but this equation isn't an identity.

As a counterexample, if a = b = pi/6, cos(a) = 1/2, cos(b) = 1/2, cos(a+b) = sqrt(3)/2 and cos(a - b) = 1.

Substituting these values into cos(a+b)cos(a-b) = cos^2(b)cos^2(a), you get sqrt(3)/2 * 1 on the left side and (1/4) * (1/4) on the right.
 
  • #3
34,183
5,801
Are you sure you copied the problem down correctly?
 
  • #4
71
0
Yes I did copy it down correctly. Obviously a mistake in the question as clearly that equality doesnt hold true

Thanks for your validation!
 

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