Determining coefficients of a trigonometric function with given tangent lines I am given the following problem:

Find a function of the form $f(x)=a+b\cos{(cx)}$ that is tangent to the line $y=1$ at the point $(0,1)$ and tangent to the line $y=x+\dfrac{3}{2}-\dfrac{\pi}{4}$ at the point $\left(\dfrac{\pi}{4},\dfrac{3}{2}\right)$.

I do not know what I am doing wrong here.  We have that $f(0)=1$ so that $$a+b\cos{(0)}=a+b=1.$$  Also, we have that$$f\left(\dfrac{\pi}{4}\right)=\dfrac{3}{2}=a+b\cos{\left(\dfrac{c\pi}{4}\right)}.$$ Taking derivatives, we know that $f'(0)=0=-bc\sin{0},$ and so this is unhelpful in solving the system.  Finaly, we have that$$f'\left(\dfrac{\pi}{4}\right)=1=-bc\sin{\left(\dfrac{c\pi}{4}\right)}.$$ Therefore we have a system of three equations with three unknowns and it should be solvable:\begin{align*}
&a+b=1\\
&a+b\cos{\left(\frac{c\pi}{4}\right)}=\frac{3}{2}\\
&bc\sin{\left(\frac{c\pi}{4}\right)}=-1
\end{align*}
So my initial thought here was to look at$$f\left(\frac{\pi}{2}\right)=a+b\cos\left(\frac{2c\pi}{4}\right)=a+b\cos\left(2\cdot\frac{c\pi}{34}\right)$$ and use double angle formulas.  So 
\begin{align*}f\left(\frac{\pi}{2}\right)&=1-b+b\left[2\cos^2{\left(\frac{c\pi}{4}\right)-1}\right]\\&=1-2b+2b\cos{\left(\frac{c\pi}{4}\right)}\cos{\left(\frac{c\pi}{4}\right)}\\&=1-2b+\frac{2}{b}\left(\frac{3}{2}-a\right)^2\\&=1-2b+\frac{2}{b}\left(\frac{1}{2}+b\right)^2\\&=1-2b+\frac{2}{b}\left(\frac{1}{4}+b+b^2\right)\\&=1-2b+\frac{1}{2b}+2+2b\\&=3+\frac{1}{2b}\end{align*}
And so this does not seem very helpful since I cannot solve for $b$ without knowing $f\left(\dfrac{\pi}{2}\right)$. I could use an alternate double angle formula, such as $\cos(2x)=1-2\sin^2{(2x)}$, and get that
$$f\left(\frac{\pi}{2}\right)=1-b+b\left[1-2\sin^2{\frac{c\pi}{4}}\right]=1-\frac{2}{bc^2}.$$
Setting them equal gives a solution for one variable, but again, this does not seem to help. I know I am missing something very basic here, but not sure what.
 A: You don't know what is $f(\pi/2)$, so indeed, it does not seem very helpful to include it in your equations.
From $a=1-b$ and the second equation, you have
$$b\left[\cos\left(\frac{c\pi}4\right)-1\right]=\frac12$$
and using now the third equation,
$$\frac{c\sin(c\pi/4)}{2\cos(c\pi/4)-2}=-1$$
or
$$c\sin(c\pi/4)+2\cos(c\pi/4)=2$$
I'd say that you need a numerical approach to solve this (equations where the unknown is "inside" and "outside" of a trascentendal function like $\sin$, $\cos$, $\exp$, etc, are not generally solvable analytically).
A: Since$$
\begin{cases}
a + b = 1\\
a + b \cos \dfrac{πc}{4} = \dfrac{3}{2}\\
-bc \sin \dfrac{πc}{4} = 1
\end{cases} \Longleftrightarrow
\begin{cases}
b \cos \dfrac{πc}{4} = b + \dfrac{1}{2}\\
b \sin \dfrac{πc}{4} = -\dfrac{1}{c}\\
a = 1 - b
\end{cases}, \tag{1}
$$
and$$
\left( b + \frac{1}{2} \right)^2 + \left( -\frac{1}{c} \right)^2 = b^2 \Longrightarrow b = -\frac{1}{c^2} - \frac{1}{4},
$$
then$$
(1) \Longleftrightarrow
\begin{cases}
\cos \dfrac{πc}{4} = -\dfrac{c^2 - 4}{c^2 + 4},\ \sin \dfrac{πc}{4} = \dfrac{4c}{c^2 + 4} & \qquad (2)\\
a = \dfrac{1}{c^2} + \dfrac{5}{4},\ b = -\dfrac{1}{c^2} - \dfrac{1}{4}
\end{cases}.
$$
WA shows that (2) may have infinitely many solutions, but solutions can only be numerically obtained.
