y=cos(x), y=Acos(Bx+C)+D Determine the values of A,B,C,and D

[Fullmetal_Hye]

y = cos(x). . .y = Acos(Bx + C) + D

(0, 1). . . . . . . .(1/2, -2)
(pi/2, 0). . . . . .(1/4, -1)
(-pi/2, 0). . . . .(-3/4, -1)

I tried like this:

. . .B(1/2) + C = 0

. . .B(1/4) + C = pi/2

. . .B(-3/4) + C = -pi/2

. . .B(1/4) + C = pi/2

. . .-B(-3/4) + C = -pi/2

. . .B = 0

But when I tried to verify that B = 0, it doesn't fit in the other equation. Where did I mess up? T.T

I need to get B in order to get the other letters.
Can you tell me how to get A and D as well?
I tried similar method but my answer won't fit in either

 

[soroban]

Re: y=cos(x), y=Acos(Bx+C)+D Determine the values of A,B,C,a

Hello, Fullmetal_Hye!

Are you sure that's the entire problem? \(\displaystyle \;\)It is strangely written.

\(\displaystyle y\:=\:\cos(x)\) . . \(\displaystyle y\:=\:A\cos(Bx\,+\,C)\,+\,D\)

\(\displaystyle \;\;\;(0,1)\) . . . . . . . . . . \(\displaystyle \left(\frac{1}{2},\,-2\right)\)

\(\displaystyle \;\;\,\left(\frac{\pi}{2},\,0\right)\) . . . . . . . . . .\(\displaystyle \left(\frac{1}{4},\,-1\right)\)

\(\displaystyle \;\;\left(-\frac{\pi}{2},\,0\right)\) . . . . . . . . . \(\displaystyle \left(-\frac{3}{4},\.-1\right)\)

Why did they include the first column?
Those facts are common knowledge ... and they don't help.


They gave us only three points for the second function.

These give us the following three equations:

\(\displaystyle \;\;-2\;=\;A\cos\left(\frac{1}{2}B\,+\,C\right)\,+\,D\)
\(\displaystyle \;\;-1\;=\;A\cos\left(\frac{1}{4}B\,+\,C\right)\,+\,D\)
\(\displaystyle \;\;-1\;= \;A\cos\left(-\frac{3}{4}B\,+\,C\right)\,+\,D\)


Are we expected to solve for \(\displaystyle A,B,C,D\) ?

 

[Fullmetal_Hye]

Nevermind, my teacher explained this to me. Thanx for your help