Question

For which value of \[\theta \] is \[\sin \theta +\sin 2\theta \] has maximum value?

Answer 1

Hint: Assume, a function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] . The slope of this function is its derivative. Use the formulas \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\] and \[\dfrac{d\left( ax \right)}{dx}=a\] , and differentiate the function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] with respect to \[\theta \] . We know that the slope at the maxima of any function is equal to zero. Now, make the derivative of the function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] equal to zero. Now, use this and get the value of \[\theta \] .

Complete step-by-step answer:
According to the question, we have the expression \[\sin \theta +\sin 2\theta \] and we have to find the value of \[\theta \] for which the expression has its maximum value.
Let us assume, a function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] …………………………(1)
The slope of a function is its derivative.
Now, differentiating the function, \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] with respect to \[\theta \] , we get
\[f'\left( \theta \right)=\dfrac{d}{d\theta }f\left( \theta \right)=\dfrac{d}{d\theta }\left( \sin \theta +\sin 2\theta \right)\]
\[\Rightarrow f'\left( \theta \right)=\dfrac{d}{d\theta }\left( \sin \theta \right)+\dfrac{d}{d\theta }\left( \sin 2\theta \right)\] ……………………….(2)
Now, using the chain rule and simplifying equation (2), we get
\[\Rightarrow f'\left( \theta \right)=\dfrac{d}{d\theta }\left( \sin \theta \right)+\dfrac{d}{d\left( 2\theta \right)}\left( \sin 2\theta \right).\dfrac{d\left( 2\theta \right)}{d\theta }\] ……………………(3)
We know the formula, \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\] and \[\dfrac{d\left( ax \right)}{dx}=a\] …………………………………(4)
Now, simplifying equation (3) by using the formulas shown in equation (4), we get
\[\Rightarrow f'\left( \theta \right)=\cos \theta +\cos 2\theta .\left( 2 \right)\]
\[\Rightarrow f'\left( \theta \right)=\cos \theta +2\cos 2\theta \]
So, the slope of the function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] is \[f'\left( \theta \right)=\cos \theta +2\cos 2\theta \] .…………………………….(5)
We know that the slope at the maxima of any function is equal to zero.
From equation (5), we have the slope of the function \[f\left( \theta \right)=\sin \theta +\sin 2\theta \] .
So, for the maxima, \[f'\left( \theta \right)=\cos \theta +2\cos 2\theta =0\] ……………………(6)
Now, solving equation (6), we get
\[\Rightarrow \cos \theta +2\cos 2\theta =0\] ……………………………(7)
We know the formula, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] ………………………………(8)
From equation (7) and equation (8), we get
\[\begin{align}
  & \Rightarrow \cos \theta +2\left( 2{{\cos }^{2}}\theta -1 \right)=0 \\
 & \Rightarrow \cos \theta +4{{\cos }^{2}}\theta -2=0 \\
\end{align}\]
\[\Rightarrow 4{{\cos }^{2}}\theta +\cos \theta -2=0\] …………………………………(9)
The above equation is quadratic in \[\cos \theta \] .
We know that the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] is \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] …………………………………(10)
Now, on comparing equation (9) and \[a{{x}^{2}}+bx+c=0\] , we get
 a = 4 ……………………………..(11)
b = 1 ……………………………(12)
c = -2 ……………………………………(13)
Now, from equation (10), equation (11), equation (12), and equation (13), we get
\[\begin{align}
  & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{{{1}^{2}}-4.4.\left( -2 \right)}}{2\left( 4 \right)} \\
 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{1+32}}{8} \\
 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{33}}{8} \\
\end{align}\]
So, \[\cos \theta =\dfrac{-1+\sqrt{33}}{8}\,=0.59307033082\] or \[\cos \theta =\dfrac{-1-\sqrt{33}}{8}\,=-0.84307033082\] .
Therefore, \[\theta \,={{\cos }^{-1}}\left( 0.59307033082 \right)\] or \[\theta ={{\cos }^{-1}}\left( -0.84307033082 \right)\] ……………………….(14)
We know that \[53.6248{}^\circ \,={{\cos }^{-1}}\left( 0.59307033082 \right)\] and \[147.4658{}^\circ ={{\cos }^{-1}}\left( -0.84307033082 \right)\] ………………………………………(15)
Now, from equation (14) and equation (15), we can say that \[\theta =53.6248{}^\circ \] or \[\theta =147.4658{}^\circ \] .

Note: In this question, one might think to use the formula \[\cos 2\theta =1-2{{\sin }^{2}}\theta \] while solving the equation \[\cos \theta +2\cos 2\theta =0\] . When we apply this formula then the equation will become a quadratic equation and there would be one cosine term also. Like,
\[\begin{align}
  & \Rightarrow \cos \theta +2\left( 1-2si{{n}^{2}}\theta \right)=0 \\
 & \Rightarrow \cos \theta -4si{{n}^{2}}\theta +2=0 \\
\end{align}\]
The above equation is not purely quadratic in \[\sin \theta \] . The presence of the one cosine term can increase the complexity of further calculations.
So, for simple calculations, we should use the formula, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\] .