Question

How do you write the trigonometric expression \[\cos \left( \arccos x+\arcsin x \right)\] as an algebraic expression.

Answer 1

Hint: In this problem, we have to write the given trigonometric expression as an algebraic expression. We know that to solve these types of problems, we have to know trigonometric formulas and identities. We can first express the given equation using a trigonometric formula. Then we can assume variables for the arc terms to convert it to an algebraic expression.

Complete step-by-step solution:
We know that the given trigonometric expression to be converted into algebraic expression is,
\[\cos \left( \arccos x+\arcsin x \right)\]…… (1)
We also know that the trigonometric identity based on this problem is,
\[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\]
Now we can apply the trigonometric formula in the expression (1), we get
\[\Rightarrow \cos \left( \arccos x+\arcsin x \right)=\cos \left( \arccos x \right)\cos \left( \arcsin x \right)-\sin \left( \arccos x \right)\sin \left( \arcsin x \right)\] ….. (2)
We know that,
\[\cos \left( \arccos x \right)=x\] …… (3)
 \[\sin \left( \arcsin x \right)=x\] ……. (4)
Now we have to find \[\cos \left( \arcsin x \right)\] and \[\sin \left( \arccos x \right)\].
We can assume that,
\[\begin{align}
  & {{\theta }_{1}}=\arcsin x \\
 & \Rightarrow \sin {{\theta }_{1}}=x \\
\end{align}\]
 Now we can use the trigonometric formula,
\[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]
We can replace the above formula with theta,
\[\begin{align}
  & \Rightarrow \cos {{\theta }_{1}}=\sqrt{1-{{\sin }^{2}}{{\theta }_{1}}}=\sqrt{1-{{x}^{2}}}\text{ }\because \text{sin}{{\theta }_{1}}=x \\
 & \Rightarrow \cos \left( \arcsin x \right)=\sqrt{1-{{x}^{2}}}....(5) \\
\end{align}\]
We can assume that,
\[\begin{align}
  & {{\theta }_{2}}=\arccos x \\
 & \Rightarrow \cos {{\theta }_{2}}=x \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \sin {{\theta }_{2}}=\sqrt{1-\cos {{\theta }_{2}}}=\sqrt{1-{{x}^{2}}} \\
 & \Rightarrow \sin \left( \arccos x \right)=\sqrt{1-{{x}^{2}}}.....(6) \\
\end{align}\]
 Now we can substitute the value (3), (4), (5), (6) in the trigonometric formula (2), we get
\[\Rightarrow \cos \left( \arccos x+\arcsin x \right)=\cos \left( \arccos x \right)\cos \left( \arcsin x \right)-\sin \left( \arccos x \right)\sin \left( \arcsin x \right)\]
\[\Rightarrow \cos \left( \arccos x+\arcsin x \right)=x\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{x}^{2}}}\] for \[\left| x \right|\le 1\].
Therefore, the algebraic expression of the trigonometric expression \[\cos \left( \arccos x+\arcsin x \right)\] is \[x\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{x}^{2}}}\] for \[\left| x \right|\le 1\].

Note: Students make mistakes while assuming values for arc terms and to find the \[\cos \left( \arcsin x \right)\]and \[\sin \left( \arccos x \right)\]. We have to use trigonometric formula \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\] to find the value of \[\cos \left( \arcsin x \right)\] and \[\sin \left( \arccos x \right)\]. We should know some trigonometric identities, formulas, properties to solve or convert the trigonometric expression to an algebraic expression.