Question

How do you verify ${{\left( \sin (\theta )+\cos (\theta ) \right)}^{2}}+{{\left( \sin (\theta )-\cos (\theta ) \right)}^{2}}=2$?

Answer 1

The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Trigonometric identities are the equations involving the trigonometric functions that are true for every value of the variables involved. These identities are true for right angled triangles. So, the Pythagorean identity of sine function is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step by step answer:
As per the given question, we need to prove the given trigonometric expression using trigonometric identities and algebraic formulae. Here, we are given the expression ${{\left( \sin (\theta )+\cos (\theta ) \right)}^{2}}+{{\left( \sin (\theta )-\cos (\theta ) \right)}^{2}}=2$ which we need to verify whether it is correct or not by solving one side of the equation.

In the given expression, if we assume $\sin \theta$ to be $a$ and $\cos \theta$ to be $b$, then we get the expression as

& \Rightarrow {{\left( \sin (\theta )+\cos (\theta ) \right)}^{2}}+{{\left( \sin (\theta )-\cos (\theta ) \right)}^{2}} \\\ & \Rightarrow {{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}} \\\ \end{aligned}$$ As we know the algebraic formulae $${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$$ and $${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$$, we can rewrite the above equation as $$\Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)+\left( {{a}^{2}}+{{b}^{2}}-2ab \right)$$ Since we have 1 outside the parentheses, we can open the brackets to get $$\Rightarrow {{a}^{2}}+{{b}^{2}}+2ab+{{a}^{2}}+{{b}^{2}}-2ab$$ Here, addition of $${{a}^{2}}$$ with $${{a}^{2}}$$ gives $$2{{a}^{2}}$$ and that of $${{b}^{2}}$$ with $${{b}^{2}}$$ gives $$2{{b}^{2}}$$. And, on subtraction of $$2ab$$ from $$2ab$$, we get zero. We can substitute these values into the previous equation to get $$\Rightarrow 2{{a}^{2}}+2{{b}^{2}}\to 2({{a}^{2}}+{{b}^{2}})$$ Now, we substitute the actual values of $$a$$ and $$b$$, we get $$\Rightarrow 2\left( {{\sin }^{2}}(\theta )+{{\cos }^{2}}(\theta ) \right)$$ Then, we use the trigonometric identity $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$ in the equation. Thus, we get $$\Rightarrow 2\left( {{\sin }^{2}}(\theta )+{{\cos }^{2}}(\theta ) \right)\to 2(1)=2$$ $$\therefore $$ This means that $${{\left( \sin (\theta )+\cos (\theta ) \right)}^{2}}+{{\left( \sin (\theta )-\cos (\theta ) \right)}^{2}}=2$$. Both the left-hand side and right-hand side are equal. Hence verified. **Note:** In order to solve such types of questions, we need to have enough knowledge over trigonometric functions and identities. We also need to know the algebraic formulae to simplify the expressions. We must avoid calculation mistakes to get the expected answers.