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

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

Explanation

Solution

We use the identity of square of addition of two numbers (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab to open the value of (sin⁡θ+cos⁡θ)2{(\sin \theta + \cos \theta )^2}. Use the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 and solve the equation into simpler form. In the end substitute the value of 2sin⁡θcos⁡θ=sin⁡2θ2\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab

Complete step-by-step answer:
We have to verify the identity (sin⁡θ+cos⁡θ)2−1=sin⁡2θ{(\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
⇒(sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ\Rightarrow {(\sin \theta + \cos \theta )^2} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta
Now we substitute the value of sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1on right hand side of the equation
⇒(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ\Rightarrow {(\sin \theta + \cos \theta )^2} = 1 + 2\sin \theta \cos \theta
Now we substitute the value of 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta on right hand side of the equation
⇒(sin⁡θ+cos⁡θ)2=1+sin⁡2θ\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
⇒(sin⁡θ+cos⁡θ)2−1=sin⁡2θ\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⁡θ+cos⁡θ)2−1=sin⁡2θ{(\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab.
⇒(sin⁡θ+cos⁡θ)2−1=(sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ)−1\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 on right hand side of the equation
⇒(sin⁡θ+cos⁡θ)2−1=1+2sin⁡θcos⁡θ−1\Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 1 + 2\sin \theta \cos \theta - 1
⇒(sin⁡θ+cos⁡θ)2−1=(1−1)+2sin⁡θcos⁡θ\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
⇒(sin⁡θ+cos⁡θ)2−1=0+2sin⁡θcos⁡θ\Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 0 + 2\sin \theta \cos \theta
⇒(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θ\Rightarrow {(\sin \theta + \cos \theta )^2} - 1 = 2\sin \theta \cos \theta
Substitute the value of 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta on right hand side of the equation
⇒(sin⁡θ+cos⁡θ)2−1=sin⁡2θ\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