Student Question

Please help ! f(x)=1-cosRx/1+cosQx        x=П/4

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The request of the problem is not specified, hence, supposing that you need to evaluate `f(x)`  at `x = pi/4` , you need to substitute pi/4 for x in equation of function, such that:

`f(pi/4) = (1-cos(R*pi/4))/(1+cos(Q*pi/4))`

Since there exists no equation to relate the coefficients R and Q, it is no possible to perform the evaluation at `x = pi/4` .

If the problem requests to find the tangent to the graph of function at the point x = pi/4, you need to use the followomg equation, such that:

`y - f(pi/4) = f'(pi/4)(x - pi/4)`

You need to find the derivative of the function using the quotient rule, such that:

`f'(x) = ((1 - cos Rx)'(1 + cos Qx) - (1 - cos Rx)(1 + cos Qx) ')/((1 + cos Qx)^2)`

`f'(x) = (Rsin Rx(1 + cos Qx) + Qsin Qx(1 - cos Rx))/((1 + cos Qx)^2)`

You need to substitute `pi/4`  for `x`  such that:

`f'(pi/4) = (Rsin Rpi/4(1 + cos Qpi/4) + Qsin Qpi/4(1 - cos Rpi/4))/((1 + cos Qpi/4)^2)`

Hence, evaluating the equation of tangent line yields `y - (1-cos(R*pi/4))/(1+cos(Q*pi/4)) = (Rsin Rpi/4(1 + cos Qpi/4) + Qsin Qpi/4(1 - cos Rpi/4))/((1 + cos Qpi/4)^2)(x - pi/4).`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial