Question

Prove that ${{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}(x),-1\le x\le 1$

Answer 1

Hint: In this question we will suppose $\cos A=x$ and we know that in the second quadrant the cosine value is negative so $\cos \left( \pi -A \right)=-\cos A=-x$. Also we will use the property of inverse trigonometric function which is ${{\cos }^{-1}}\cos y=y$, where $0\le y\le \pi$.

Complete step-by-step answer:
We have been given to prove:
${{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}(x),-1\le x\le 1$
Let us suppose $\cos A=x$
Since we know that in the second quadrant the cosine value is negative.
So, $\cos \left( \pi -A \right)=-\cos A=-x$
We have $\cos A=x$
Taking inverse cosine function to both side of equation to get,
$co{{s}^{-1}}\cos A={{\cos }^{-1}}x$
Since $co{{s}^{-1}}\operatorname{cosy}={{y}_{1}}$, where $0\le y\le \pi$

& A={{\cos }^{-1}}x \\\ & \Rightarrow {{\cos }^{-1}}x=A......(1) \\\ \end{aligned}$$ Also we have $$\cos \left( \pi -A \right)=-x$$ Taking inverse cosine function to both the sides of the equality, we get as follows: $${{\cos }^{-1}}\cos \left( \pi -A \right)={{\cos }^{-1}}(-x)$$ Since we know that $${{\cos }^{-1}}\cos y=y$$, where $$0\le y\le \pi $$ $$\begin{aligned} & \Rightarrow \pi -A={{\cos }^{-1}}(-x) \\\ & \Rightarrow {{\cos }^{-1}}(-x)=\pi -A.....(2) \\\ \end{aligned}$$ Adding equation (1) and (2) we get as follows: $$\begin{aligned} & {{\cos }^{-1}}x+{{\cos }^{-1}}(-x)=\pi -A+A \\\ & \Rightarrow {{\cos }^{-1}}x+{{\cos }^{-1}}(-x)=\pi \\\ \end{aligned}$$ On subtracting $${{\cos }^{-1}}x$$ from both the sides of the equation we get as follows: $$\begin{aligned} & {{\cos }^{-1}}-{{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}(x) \\\ & \Rightarrow {{\cos }^{-1}}(-x)=\pi -{{\cos }^{-1}}(x) \\\ \end{aligned}$$ Hence Proved. Note: Be careful of the sign mistakes during the calculation. Also remember that the value of ‘A’ must be less than or equal to $$\dfrac{\pi }{2}$$ radians otherwise we can’t say that $$\cos \left( \pi -A \right)=-\cos A$$. Also, remember that in the first quadrant all trigonometric ratios are positive, in the second quadrant sine and cosecant are positive, in third quadrant tangent and cotangent and in fourth quadrant cosine and secant are positive.