Question

If a trigonometric expression is given as \[\sin A+\cos A=1,\]then \[\sin 2A\] is equal to.
(A). 1
(B). 2
(C). 0
(D). \[\dfrac{1}{2}\]

Answer 1

Hint: Square both sides of the given equation. And use the identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify the expression after applying square to both the sides. Now, replace \[2\sin A\cos A\] by \[\sin 2A\] using the relation.

Complete step-by-step solution -
\[\sin 2\theta =2\sin \theta \cos \theta \]
And for squaring both the sides, use
\[({{a}^{2}}+{{b}^{2}})={{a}^{2}}+{{b}^{2}}+2ab\]
Here, it is given that
\[\sin A+\cos A=1\] …….. (i)
And hence, we need to find value of \[\sin 2A\].
As we know trigonometric identity of \[\sin 2A\] is given as
\[\sin 2A=2\sin A\cos A\] ……. (ii)
And we also know the algebraic identity of \[{{(a+b)}^{2}}\] is given as
\[({{a}+{b}}^{2})={{a}^{2}}+{{b}^{2}}+2ab\] ……… (iii)
Now, we can observe that if we square the equation (i) to both sides and observe the L.H.S with the help of identity (iii), we get the equation (i) as
\[\begin{align}
  & {{(\sin A+\cos A)}^{2}}={{(1)}^{2}} \\
  \Rightarrow & {{(\sin A)}^{2}}+{{(\cos A)}^{2}}+2\sin A\cos A=1 \\
   \Rightarrow & {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=1........(iv) \\
\end{align}\]
Now, as we know the trigonometric identity of \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta \] is given as
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] ……… (v)
Hence, we can replace \[{{\sin }^{2}}A+{{\cos }^{2}}A\] by 1 in the equation (iv) using equation (v). So, we can re-write equation (iv) as
\[1+\sin 2A-1=1-1\]
Or
\[\sin 2A=0\]
Hence, we get the value of \[\sin 2A\] as 0, if \[\sin 2A+\cos A=1\].
So, 0 is the answer to the problem.

Note: Another approach to solve the question would be
\[\begin{align}
  & \cos A=1-\sin A \\
 & {{\cos }^{2}}A={{(1-\sin A)}^{2}} \\
\end{align}\]
Now, put $\cos^{2}A = 1- \sin^{2}A $ as we know the relation
\[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]
So, we get \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]
\[\begin{align}
  \Rightarrow & 1-{{\sin }^{2}}A=1+{{\sin }^{2}}A-2\sin A \\
   \Rightarrow & 2{{\sin }^{2}}A-2\sin A=0 \\
   \Rightarrow & \sin A(\sin A-1)=0 \\
\end{align}\]
\[\sin A=0\] Or \[\sin A=1\]
i.e. \[A={{0}^{o}}\] or \[A=\dfrac{\pi }{2}\]
Now, use \[\sin 2A=2\sin A\cos A\] to get the value of \[\sin 2A\] as 0. So, it can be another approach.
One may directly guess the angle A by observing the equation \[\sin A+\cos A=1\]. As a sum of \[\sin A\] and \[\cos A\] can only be 1, if A=0 or \[\dfrac{\pi }{2}\]. i.e. if \[\sin A=0\], then \[\cos A\] should be 1 or vice versa. Hence, one can answer the value of \[\sin 2A\] directly as well.