Solveeit Logo
Login

Question

Question: By using the suitable trigonometric identities and formulas expand \({{\left( \cos \theta -\sin \the...

By using the suitable trigonometric identities and formulas expand (cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2=.........{{\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) 2cos⁡2θ2{{\cos }^{2}}\theta
(d) 2sin⁡2θ2{{\sin }^{2}}\theta

Explanation

Solution

Hint : For solving this problem, first we try to simplify the given expression by expanding (cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2{{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}} with the help of algebraic expansion (a+b)2=a2+b2+2ab and (a−b)2=a2+b2−2ab{{\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⁡2θ+sin⁡2θ=1{{\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 (cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2…(1){{\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 (a+b)2=a2+b2+2ab and (a−b)2=a2+b2−2ab{{\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
⇒cos⁡2θ+sin⁡2θ−2cos⁡θsin⁡θ+cos⁡2θ+sin⁡2θ+2cos⁡θsin⁡θ\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\cos \theta \sin \theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta
−2sin⁡θcos⁡θ and 2sin⁡θcos⁡θ-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:
⇒cos⁡2θ+sin⁡2θ+cos⁡2θ+sin⁡2θ\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta
Adding the like terms in the expression, we get:
⇒2cos⁡2θ+2sin⁡2θ\Rightarrow 2{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta
Taking 2 common, we get:
⇒2(cos⁡2θ+sin⁡2θ)\Rightarrow 2\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)
As we know that the value of trigonometry identity (cos⁡2θ+sin⁡2θ)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) is 1. By using this we get:
⇒2×1=2 ∴(cos⁡θ−sin⁡θ)2+(cos⁡θ+sin⁡θ)2=2 \begin{aligned} & \Rightarrow 2\times 1=2 \\\ & \therefore {{\left( \cos \theta -\sin \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}=2 \\\ \end{aligned}
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⁡θ−sin⁡θ=a and cos⁡θ+sin⁡θ=b\cos \theta -\sin \theta =a\text{ and }\cos \theta +\sin \theta =b . Now by using the algebraic identity, it can be simplified as:
⇒a2+b2=(a+b)2−2ab ⇒(a+b)=2cos⁡θ and 2ab=2(cos⁡2θ−sin⁡2θ) \begin{aligned} & \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{aligned}
By replacing these values in the above formula, we obtain the same result as obtained above.