
Prove that $\cos \left( {{\cos }^{-1}}x \right)=x,x\in \left[ -1,1 \right]$
Answer
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Hint: Use the fact that if $y={{\cos }^{-1}}x$, then $x=\cos y$. Assume $y={{\cos }^{-1}}x$. Write $\cos \left( {{\cos }^{-1}}x \right)$ in terms of y and hence prove the above result.
Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how ${{\cos }^{-1}}x$ is defined even when $\cos x$ is not one-one.
We know that cosx is a periodic function.
Let us draw the graph of cosx
As is evident from the graph cosx is a repeated chunk of the graph of cosx within the interval [A, B] and it attains all its possible values in the interval [A, C]. Here $A=0,B=2\pi $ and $C=\pi $.
Hence if we consider cosx in the interval [A, C], we will lose no value attained by cosx, and at the same time, cosx will be one-one and onto.
Hence ${{\cos }^{-1}}x$ is defined over the Domain $\left[ -1,1 \right]$, with codomain $\left[ 0,\pi \right]$ as in the Domain $\left[ 0,\pi \right]$, cosx is one-one and ${{R}_{\cos x}}=\left[ -1,1 \right]$.
Now since ${{\cos }^{-1}}x$ is the inverse of cosx it satisfies the fact that if $y={{\cos }^{-1}}x$, then $\cos y=x$.
So let $y={{\cos }^{-1}}x$
Hence we have cosy = x.
Now $\cos \left( {{\cos }^{-1}}x \right)=\cos y$
Hence we have $\cos \left( {{\cos }^{-1}}x \right)=x$.
Also as x is the Domain of ${{\cos }^{-1}}x$, we have $x\in \left[ -1,1 \right]$.
Hence $\cos \left( {{\cos }^{-1}}x \right)=x,x\in \left[ -1,1 \right]$
Note: [1] The above-specified codomain for ${{\cos }^{-1}}x$ is called principal branch for ${{\cos }^{-1}}x$. We can select any branch as long as $\cos x$ is one-one and onto and Range $=\left[ -1,1 \right]$. The proof will remain the same as above.
Complete step-by-step answer:
Before dwelling into the proof of the above question, we must understand how ${{\cos }^{-1}}x$ is defined even when $\cos x$ is not one-one.
We know that cosx is a periodic function.
Let us draw the graph of cosx
As is evident from the graph cosx is a repeated chunk of the graph of cosx within the interval [A, B] and it attains all its possible values in the interval [A, C]. Here $A=0,B=2\pi $ and $C=\pi $.
Hence if we consider cosx in the interval [A, C], we will lose no value attained by cosx, and at the same time, cosx will be one-one and onto.
Hence ${{\cos }^{-1}}x$ is defined over the Domain $\left[ -1,1 \right]$, with codomain $\left[ 0,\pi \right]$ as in the Domain $\left[ 0,\pi \right]$, cosx is one-one and ${{R}_{\cos x}}=\left[ -1,1 \right]$.
Now since ${{\cos }^{-1}}x$ is the inverse of cosx it satisfies the fact that if $y={{\cos }^{-1}}x$, then $\cos y=x$.
So let $y={{\cos }^{-1}}x$
Hence we have cosy = x.
Now $\cos \left( {{\cos }^{-1}}x \right)=\cos y$
Hence we have $\cos \left( {{\cos }^{-1}}x \right)=x$.
Also as x is the Domain of ${{\cos }^{-1}}x$, we have $x\in \left[ -1,1 \right]$.
Hence $\cos \left( {{\cos }^{-1}}x \right)=x,x\in \left[ -1,1 \right]$
Note: [1] The above-specified codomain for ${{\cos }^{-1}}x$ is called principal branch for ${{\cos }^{-1}}x$. We can select any branch as long as $\cos x$ is one-one and onto and Range $=\left[ -1,1 \right]$. The proof will remain the same as above.
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