Given,$\sin 2A=2\sin A$ is true when A is equal to:
A. 0
B. 1
C. 4
D. -1
Answer
596.7k+ views
Hint: To solve this question we will use the trigonometric formula of $\sin 2A=2\sin A\cos A$. We will solve the equation by using the formula and find the value of A to get the desired answer. Trigonometric functions and values will be used to solve this question.
Complete answer:
We have been given an expression $\sin 2A=2\sin A$.
We have to find the value of A for which the given expression is true.
To solve this question first we will take the terms to the left side, we get
$\Rightarrow \sin 2A-2\sin A=0$
Now, we know that we have a trigonometric formula $\sin 2A=2\sin A\cos A$
So, substituting the value of the formula in the given equation we get
$\Rightarrow 2\sin A\cos A-2\sin A=0$
Taking the common terms out we get
$\Rightarrow 2\sin A\left( \cos A-1 \right)=0$
Or we can write the above equation as
$\Rightarrow 2\sin A=0\text{ and }\left( \cos A-1 \right)=0$
Now, we have
$\begin{align}
& \Rightarrow 2\sin A=0\text{ and }\left( \cos A-1 \right)=0 \\
& \Rightarrow \sin A=0\text{ and }\cos A=1 \\
\end{align}$
Now, we know that $\sin 0{}^\circ =0\text{ and cos0}{}^\circ \text{=1}$
So, we get $\Rightarrow \sin A=\sin 0{}^\circ \text{ and }\cos A=\cos 0{}^\circ $
So, we get $A=0$
So, $\sin 2A=2\sin A$ is true when A is equal to $0$.
Option A is the correct answer.
Note:
Alternative way to solve this question is as follows:
We know that $\sin 2A=2\sin A\cos A$
For $\sin 2A=2\sin A$ to be true the value of $\cos A$ must be equal to 1.
Then, we know that $\text{cos0}{}^\circ \text{=1}$.
So, $\sin 2A=2\sin A$ to be true the value of A must be equal to 0.
To solve these types of questions students must have a knowledge of trigonometric functions and identities.
Complete answer:
We have been given an expression $\sin 2A=2\sin A$.
We have to find the value of A for which the given expression is true.
To solve this question first we will take the terms to the left side, we get
$\Rightarrow \sin 2A-2\sin A=0$
Now, we know that we have a trigonometric formula $\sin 2A=2\sin A\cos A$
So, substituting the value of the formula in the given equation we get
$\Rightarrow 2\sin A\cos A-2\sin A=0$
Taking the common terms out we get
$\Rightarrow 2\sin A\left( \cos A-1 \right)=0$
Or we can write the above equation as
$\Rightarrow 2\sin A=0\text{ and }\left( \cos A-1 \right)=0$
Now, we have
$\begin{align}
& \Rightarrow 2\sin A=0\text{ and }\left( \cos A-1 \right)=0 \\
& \Rightarrow \sin A=0\text{ and }\cos A=1 \\
\end{align}$
Now, we know that $\sin 0{}^\circ =0\text{ and cos0}{}^\circ \text{=1}$
So, we get $\Rightarrow \sin A=\sin 0{}^\circ \text{ and }\cos A=\cos 0{}^\circ $
So, we get $A=0$
So, $\sin 2A=2\sin A$ is true when A is equal to $0$.
Option A is the correct answer.
Note:
Alternative way to solve this question is as follows:
We know that $\sin 2A=2\sin A\cos A$
For $\sin 2A=2\sin A$ to be true the value of $\cos A$ must be equal to 1.
Then, we know that $\text{cos0}{}^\circ \text{=1}$.
So, $\sin 2A=2\sin A$ to be true the value of A must be equal to 0.
To solve these types of questions students must have a knowledge of trigonometric functions and identities.
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