Learn inverse trigonometric functions & sine, cosine, tangent functions in calculus. Find simplification of expression containing a trigonometric and an inverse trigonometric function.
If we apply the horizontal line test we see that \(f\) is not one to one.
But if we restrict the domain of \(sin\,x\) to \(\left [ \dfrac {-\pi}{2},\dfrac {\pi}{2} \right]\)then graph is
Now the function becomes one-to-one.
We say that the inverse function of the restricted sine function is \(sin^{-1}x\) or arc \((sin\,x)\)
\(\Rightarrow sin^{-1}\left (-\dfrac {1}{2}\right)=\dfrac {-\pi}{6}\)
A \( \dfrac {\pi}{6}\)
B \( \dfrac {-\pi}{3}\)
C \( \dfrac {-\pi}{6}\)
D \( \dfrac {\pi}{3}\)
\(\therefore\;cos^{-1}\Big(-\dfrac {1}{2}\Big)=\dfrac {2\pi}{3}\)
A \(\dfrac {\pi}{6}\)
B \(\dfrac {-\pi}{6}\)
C \(\dfrac {\pi}{3}\)
D \(\dfrac {-\pi}{3}\)
The above formulae are a direct result of the fact that \(fof^{-1}(x)=x\) for all valid \(x\).
A \(\dfrac {\pi}{6}\)
B \(\dfrac {\pi}{3}\)
C \(\dfrac {-\pi}{3}\)
D \(\dfrac {\pi}{2}\)
e.g Consider the expression
\(tan (sin^{-1}x)\)
Let \(sin^{-1} x=y \Rightarrow x=sin \,y\) where \(y\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)
Now if \(sin \,y = x \Rightarrow cos\,y = \sqrt{1-x^2}\)
\(\Rightarrow sec\,y =\dfrac{1}{\sqrt{1-x^2}} \Rightarrow tan\,y \sqrt{\dfrac{1}{1-x^2}-1} = \sqrt{\dfrac{x^2}{1-x^2}}\)
\(\Rightarrow tan\,y = \dfrac{x}{\sqrt{1-x^2}} \to\) which is the required expression
A \(\dfrac {x}{\sqrt{1-x^2}}\)
B \(\dfrac {x}{\sqrt{1+x^2}}\)
C \(\dfrac {\sqrt{1+x^2}}{x}\)
D \(\dfrac {\sqrt{1-x^2}}{x}\)
The inverse function of this restricted tangent function is called inverse tangent function or \(tan^{-1}\,x\) or arc\((tan\,x)\).
\(\therefore \;\; tan^{-1}(-1)=\dfrac {-\pi}{4}\)
A \(\dfrac {\pi}{ 3}\)
B \(\dfrac {-\pi}{ 6}\)
C \(\dfrac {-\pi}{ 3}\)
D \(\dfrac {-\pi}{ 2} \)
The above formulae are a direct result of the fact that \(fof^{-1}(x)=x\) for all valid \(x\).
A \(\dfrac {7\,\pi}{6}\)
B \(\dfrac {\pi}{6}\)
C \(\dfrac {13\,\pi}{6}\)
D \(\dfrac {\pi}{4}\)