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.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...
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.
To use this equation, you need to know the values of a and b. Plug these values into the equation to calculate da/dx.
Yes, the equation can be simplified to da/dx = tan(b - a), which may be easier to use in certain situations.
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.
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.