Finding the Derivative of x = sin(a)/cos(b) with Given Constraints

  • Thread starter Gekko
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In summary, the conversation discusses an attempt to show that cos(a+b)cos(a-b) is equal to cos^2(b)cos^2(a). However, this equation is not an identity as shown through a counterexample provided. The person asking for help suspects there may be a mistake in the original question.
  • #1
Gekko
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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...
 
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  • #2
Gekko said:

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
Are you sure you copied the problem down correctly?
 
  • #4
Yes I did copy it down correctly. Obviously a mistake in the question as clearly that equality doesn't hold true

Thanks for your validation!
 

FAQ: Finding the Derivative of x = sin(a)/cos(b) with Given Constraints

What is the equation for finding da/dx in terms of sin and cos?

The equation for finding da/dx in terms of sin and cos is da/dx = (cos(a)cos(b) - sin(a)sin(b)) / (cos(b))^2.

How do I use this equation to find da/dx?

To use this equation, you need to know the values of a and b. Plug these values into the equation to calculate da/dx.

Can I simplify the equation to make it easier to use?

Yes, the equation can be simplified to da/dx = tan(b - a), which may be easier to use in certain situations.

What does the value of da/dx represent?

The value of da/dx represents the rate of change of the angle a with respect to the variable x. It is often used in differential calculus to find the instantaneous rate of change at a specific point on a curve.

Are there any other methods for finding da/dx?

Yes, there are other methods such as using the quotient rule or the chain rule in calculus. However, for this specific equation, the da/dx = (cos(a)cos(b) - sin(a)sin(b)) / (cos(b))^2 equation is the most straightforward and efficient method.

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