Trig and Inverse Trig Graphs

[diasy77]

The graphs of the function f(x) = cos(2x) and f-1(x)=cos-1(2x) intersect at x=a (a>0). Explain why a is a solution of the equation x-cos-1(2x)=0

 

[Deleted member 4993]

The graphs of the function f(x) = cos(2x) and f-1(x)=cos-1(2x) intersect at x=a (a>0). Explain why a is a solution of the equation x-cos-1(2x)=0

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[stapel]

The graphs of the function f(x) = cos(2x) and f-1(x)=cos-1(2x) intersect at x=a (a>0). Explain why a is a solution of the equation x-cos-1(2x)=0

If they "intersect", then what point is on both lines? How does this relate to "x = a", "y = cos(2a)", and "y^-1 = cos^-1(2a)"?

 

[Ishuda]

The graphs of the function f(x) = cos(2x) and f-1(x)=cos-1(2x) intersect at x=a (a>0). Explain why a is a solution of the equation x-cos-1(2x)=0

If you mean that f-1(x) is the inverse function of f(x)=cos(2x), i.e. f(f-1(x))=x, then the arccosine of 2x is not that inverse function. That is
\(\displaystyle \text{If }f(x)=cos(2x)\, then\, f^{-1}(x)\ne\,cos^{-1}(2x)\)