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

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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).` **

Thanks !!)