Question

By using the suitable trigonometric identities and formulas expand ${{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}=.........$ to the most reducible form having only one trigonometric term if required or to the simple integral form. Select the appropriate correct alternatives from the below available options after obtaining the final equivalence of the given expression.
(a) 0
(b) 2
(c) $2{{\cos }^{2}}\theta $
(d) $2{{\sin }^{2}}\theta $

Answer 1

Hint: For solving this problem, first we try to simplify the given expression by expanding ${{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}$ with the help of algebraic expansion ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\text{ and }{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Now, we rearrange the terms to get the common part separated from the expression. Finally, by applying the trigonometric identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$, we get the desired result.

Complete step by step solution:
According to the problem statement, we are given that ${{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}\ldots \left( 1 \right)$
We are required to evaluate the final simplified expression which can be numerical, or expression as stated in options.
First, we simplified the above expression using the algebraic identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\text{ and }{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Now, employing these identities in equation (1) and replacing a as cosine of the angle and b as sin of the angle, we get
$\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\cos \theta \sin \theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta $
$-2\sin \theta \cos \theta \text{ and }2\sin \theta \cos \theta $ have opposite sign, so they cancel out each other. Now, the simplified expression can be expressed as:
$\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta $
Adding the like terms in the expression, we get:
$\Rightarrow 2{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta $
Taking 2 common, we get:
$\Rightarrow 2\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)$
As we know that the value of trigonometry identity $\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)$ is 1. By using this we get:
$\begin{align}
  & \Rightarrow 2\times 1=2 \\
 & \therefore {{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}=2 \\
\end{align}$
Therefore, the simplified form of the function stated in problem is 2.
Hence, option (b) is correct.

Note: This problem can be alternatively solved by using substitution of $\cos \theta -\sin \theta =a\text{ and }\cos \theta +\sin \theta =b$ . Now by using the algebraic identity, it can be simplified as:
$\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\
 & \Rightarrow \left( a+b \right)=2\cos \theta \ and\ 2ab=2\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right) \\
\end{align}$
By replacing these values in the above formula, we obtain the same result as obtained above.