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

& \Rightarrow \cos \theta +2\left( 2{{\cos }^{2}}\theta -1 \right)=0 \\\ & \Rightarrow \cos \theta +4{{\cos }^{2}}\theta -2=0 \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ So, $$\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{aligned} & \Rightarrow \cos \theta +2\left( 1-2si{{n}^{2}}\theta \right)=0 \\\ & \Rightarrow \cos \theta -4si{{n}^{2}}\theta +2=0 \\\ \end{aligned}$$ The above equation is not purely quadratic in $$\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$$ .