
The value of $\theta $ satisfying $\sin 7\theta =\sin 4\theta -\sin \theta $ and $0<\theta <\dfrac{\pi }{2}$, are
A. \[\dfrac{\pi }{9},\dfrac{\pi }{4}\]
B. \[\dfrac{\pi }{3},\dfrac{\pi }{9}\]
C. \[\dfrac{\pi }{6},\dfrac{\pi }{9}\]
D. \[\dfrac{\pi }{3},\dfrac{\pi }{4}\]
Answer
163.8k+ views
Hint:To find the value of $\theta $ we will simplify the given equation using the trigonometric formula $\sin C+\sin D$. After simplifying the equation, we will get two factors which we will equate to zero. Using trigonometric table of values, we will derive the value of $\theta $. Then we will see which value of $\theta $ lies in the given interval.
Formula Used: $\sin C+\sin D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
Complete step by step solution: We are given a trigonometric equation $\sin 7\theta =\sin 4\theta -\sin \theta $ where $0<\theta <\dfrac{\pi }{2}$ and we have to derive the value of $\theta $.
We will take the given equation and apply the formula of $\sin C+\sin D$and then simplify.
\[\begin{align}
& \sin 7\theta +\sin \theta =\sin 4\theta \\
& 2\sin \dfrac{7\theta +\theta }{2}\cos \dfrac{7\theta -\theta }{2}=\sin 4\theta \\
& 2\sin \dfrac{7\theta +\theta }{2}\cos \dfrac{7\theta -\theta }{2}=\sin 4\theta \\
& 2\sin 4\theta \cos 3\theta -\sin 4\theta =0 \\
& \sin 4\theta (2\cos 3\theta -1)=0
\end{align}\]
Now we will equate both the factors to zero.
$\sin 4\theta =0$ or $\begin{align}
& 2\cos 3\theta -1=0 \\
& \cos 3\theta =\dfrac{1}{2}
\end{align}$.
We know that $\sin 0=0,\sin \pi =0$ and $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$. So,
$\begin{align}
& \sin 4\theta =\sin 0 \\
& 4\theta =0 \\
& \theta =0
\end{align}$ , $\begin{align}
& \sin 4\theta =\sin \pi \\
& 4\theta =\pi \\
& \theta =\dfrac{\pi }{4}
\end{align}$ or $\begin{align}
& \cos 3\theta =\cos \dfrac{\pi }{3} \\
& 3\theta =\dfrac{\pi }{3} \\
& \theta =\dfrac{\pi }{9}
\end{align}$.
As we are given that the value of $\theta $ lies in between the interval of $0<\theta <\dfrac{\pi }{2}$ so the value of $\theta $ will be $\theta =\dfrac{\pi }{4},\dfrac{\pi }{9}$.
The value of $\theta $ for the trigonometric equation $\sin 7\theta =\sin 4\theta -\sin \theta $ when $0<\theta <\dfrac{\pi }{2}$ is $\dfrac{\pi }{4},\dfrac{\pi }{9}$
Option ‘A’ is correct
Note: Here we directly derived the principal solution of angle instead of finding the general solution. We can also use the theorem of general solution of angle for sin that is $x=n\pi +{{(-1)}^{n}}y$ and cos that is $x=2n\pi \pm y$ and then substitute the value of $n$ and derive the possible solutions for the angle. And from the all the solutions could have selected the value of $\theta $according to the interval given.
In the question given $\sin 7\theta =\sin 4\theta -\sin \theta $ we selected the terms $\sin 7\theta $ and $\sin \theta $ to apply the formula $\sin C+\sin D$ because if we select any other terms then we will get the angle in fractions.
Formula Used: $\sin C+\sin D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
Complete step by step solution: We are given a trigonometric equation $\sin 7\theta =\sin 4\theta -\sin \theta $ where $0<\theta <\dfrac{\pi }{2}$ and we have to derive the value of $\theta $.
We will take the given equation and apply the formula of $\sin C+\sin D$and then simplify.
\[\begin{align}
& \sin 7\theta +\sin \theta =\sin 4\theta \\
& 2\sin \dfrac{7\theta +\theta }{2}\cos \dfrac{7\theta -\theta }{2}=\sin 4\theta \\
& 2\sin \dfrac{7\theta +\theta }{2}\cos \dfrac{7\theta -\theta }{2}=\sin 4\theta \\
& 2\sin 4\theta \cos 3\theta -\sin 4\theta =0 \\
& \sin 4\theta (2\cos 3\theta -1)=0
\end{align}\]
Now we will equate both the factors to zero.
$\sin 4\theta =0$ or $\begin{align}
& 2\cos 3\theta -1=0 \\
& \cos 3\theta =\dfrac{1}{2}
\end{align}$.
We know that $\sin 0=0,\sin \pi =0$ and $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$. So,
$\begin{align}
& \sin 4\theta =\sin 0 \\
& 4\theta =0 \\
& \theta =0
\end{align}$ , $\begin{align}
& \sin 4\theta =\sin \pi \\
& 4\theta =\pi \\
& \theta =\dfrac{\pi }{4}
\end{align}$ or $\begin{align}
& \cos 3\theta =\cos \dfrac{\pi }{3} \\
& 3\theta =\dfrac{\pi }{3} \\
& \theta =\dfrac{\pi }{9}
\end{align}$.
As we are given that the value of $\theta $ lies in between the interval of $0<\theta <\dfrac{\pi }{2}$ so the value of $\theta $ will be $\theta =\dfrac{\pi }{4},\dfrac{\pi }{9}$.
The value of $\theta $ for the trigonometric equation $\sin 7\theta =\sin 4\theta -\sin \theta $ when $0<\theta <\dfrac{\pi }{2}$ is $\dfrac{\pi }{4},\dfrac{\pi }{9}$
Option ‘A’ is correct
Note: Here we directly derived the principal solution of angle instead of finding the general solution. We can also use the theorem of general solution of angle for sin that is $x=n\pi +{{(-1)}^{n}}y$ and cos that is $x=2n\pi \pm y$ and then substitute the value of $n$ and derive the possible solutions for the angle. And from the all the solutions could have selected the value of $\theta $according to the interval given.
In the question given $\sin 7\theta =\sin 4\theta -\sin \theta $ we selected the terms $\sin 7\theta $ and $\sin \theta $ to apply the formula $\sin C+\sin D$ because if we select any other terms then we will get the angle in fractions.
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