Question

How do you verify the identity \[{(\sin \theta + \cos \theta )^2} - 1 = \sin 2\theta \]?

Answer 1

Hint: We use the identity of square of addition of two numbers \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] to open the value of \[{(\sin \theta + \cos \theta )^2}\]. Use the identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and solve the equation into simpler form. In the end substitute the value of \[2\sin \theta \cos \theta = \sin 2\theta \] to get the final answer.
* If ‘a’ and ‘b’ are two numbers then we can write \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]

Complete step-by-step answer:
We have to verify the identity \[{(\sin \theta + \cos \theta )^2} - 1 = \sin 2\theta \]
We will open the square of sum of sine and cosine of the angle using the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta \]
Now we substitute the value of \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]on right hand side of the equation
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} = 1 + 2\sin \theta \cos \theta \]
Now we substitute the value of \[2\sin \theta \cos \theta = \sin 2\theta \]on right hand side of the equation
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} = 1 + \sin 2\theta \]
Shift the constant value to left hand side of the equation as required in the question
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = \sin 2\theta \]
This is the required equation.

Hence Proved

Note:
Alternate method:
We can show the left hand side of the equation is equal to the right hand side by substituting the value of identity in the left hand side of the equation. Simultaneously substitute the values from trigonometric identities to solve the left side and make it equal to the right hand side. We have to show\[{(\sin \theta + \cos \theta )^2} - 1 = \sin 2\theta \].
Here we start with the left hand side. Open the identity in the bracket using the identity\[{(a + b)^2} = {a^2} + {b^2} + 2ab\].
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = ({\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta ) - 1\]
Now substitute the value of \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] on right hand side of the equation
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 1 + 2\sin \theta \cos \theta - 1\]
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = (1 - 1) + 2\sin \theta \cos \theta \]
Now add the constant values on right hand side of the equation
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 0 + 2\sin \theta \cos \theta \]
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 2\sin \theta \cos \theta \]
Substitute the value of \[2\sin \theta \cos \theta = \sin 2\theta \] on right hand side of the equation
\[ \Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = \sin 2\theta \]
This is equal to the right hand side of the equation.
So, the left hand side of the equation is equal to the right hand side of the equation.
Hence Proved