Question

If $\sin \theta + \cos \theta = 1$, then $\sin 2\theta = $
$
  A.{\text{ 0}} \\
  B.{\text{ }}\dfrac{1}{2} \\
  C.{\text{ 1}} \\
  D.{\text{ 2}} \\
$

Answer 1

Hint: In this particular question first squaring on both sides later on use some basic trigonometric identities such as \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] to reach the solution of the question.

Complete step-by-step answer:
Given trigonometric equation is

$\sin \theta + \cos \theta = 1$…………………… (1)

Then we have to find out the value of $\sin 2\theta $

Now in equation (1) squaring both sides we have,

$ \Rightarrow {\left( {\sin \theta + \cos \theta } \right)^2} = {1^2} = 1$

Now open the square according the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

we have,

\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 1\]…………….. (2)

Now as we know that the value of \[{\sin ^2}\theta + {\cos ^2}\theta \] is 1 and \[2\sin

\theta \cos \theta \] is written as $\sin 2\theta $.

So use these properties in equation (2) we have,

\[ \Rightarrow 1 + \sin 2\theta = 1\]

Now cancel out 1 from both sides we have,

\[ \Rightarrow \sin 2\theta = 0\]

So, the value of $\sin 2\theta $ is zero (0) if $\sin \theta + \cos \theta = 1$.

So this is the required answer.

Hence option (A) is correct.

Note: Whenever we face such types of questions the key concept is squaring on both sides, then open the square using the property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ then apply some basic trigonometric identities as stated above and simplify, we will get the required answer of $\sin 2\theta $.