Question

Consider the following trigonometric equation: $2\sin \left( \theta +\dfrac{\pi }{3} \right)=\cos \left( \theta -\dfrac{\pi }{3} \right)$ and $\tan \theta +\sqrt{3}=0$. Find the value of $\theta$.

Answer 1

Hint: Use the trigonometric identities of $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ and $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$ and apply it in the first equation. We will also use values of sin and cos at angles like $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$ to simplify it further. Hence find the value of $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and put it in $\tan \theta +\sqrt{3}=0$. Hence solve it and get the value of $\theta$.

Complete step-by-step answer:
We have been given a trigonometric expression, which is,
$\Rightarrow 2\sin \left( \theta +\dfrac{\pi }{3} \right)=\cos \left( \theta -\dfrac{\pi }{3} \right)$ - (1)
We know the basic trigonometric identities such as,

& \Rightarrow \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b \\\ & \Rightarrow \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b \\\ \end{aligned}$$ Now apply these properties for $$\sin \left( \theta +\dfrac{\pi }{3} \right)$$ and $$\cos \left( \theta -\dfrac{\pi }{3} \right)$$ in equation (1). Thus we can write, $$\Rightarrow 2\sin \left( \theta +\dfrac{\pi }{3} \right)=\cos \left( \theta -\dfrac{\pi }{3} \right)$$ $$\Rightarrow 2\left[ \sin \theta \cos \dfrac{\pi }{3}+\cos \theta \sin \dfrac{\pi }{3} \right]=\cos \theta \cos \dfrac{\pi }{3}+\sin \theta \sin \dfrac{\pi }{3}$$ - (2) With the help of a trigonometric table, we can find the $$\cos \dfrac{\pi }{3}$$ and $$\sin \dfrac{\pi }{3}$$. | $${{0}^{\circ }}$$| $${{30}^{\circ }}$$| $${{45}^{\circ }}$$| $${{60}^{\circ }}$$| $${{90}^{\circ }}$$ ---|---|---|---|---|--- | 0| $$\dfrac{\pi }{6}$$| $$\dfrac{\pi }{4}$$| $$\dfrac{\pi }{3}$$| $$\dfrac{\pi }{2}$$ sin| 0| $$\dfrac{1}{2}$$| $$\dfrac{1}{\sqrt{2}}$$| $$\dfrac{\sqrt{3}}{2}$$| 1 cos| 1| $$\dfrac{\sqrt{3}}{2}$$| $$\dfrac{1}{\sqrt{2}}$$| $$\dfrac{1}{2}$$| 0 tan| 0| $$\dfrac{1}{\sqrt{3}}$$| 1| $$\sqrt{3}$$| N.D Thus from the table, we get the value as, $$\cos \dfrac{\pi }{3}=\dfrac{1}{2}$$ and $$\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$$. Hence put the values in (2), $$\begin{aligned} & 2\left[ \sin \theta \times \dfrac{1}{2}+\cos \theta \times \dfrac{\sqrt{3}}{2} \right]=\cos \theta \times \dfrac{1}{2}+\sin \theta \times \dfrac{\sqrt{3}}{2} \\\ & \dfrac{2}{2}\left[ \sin \theta +\sqrt{3}\cos \theta \right]=\dfrac{1}{2}\left( \cos \theta +\sqrt{3}\sin \theta \right) \\\ & \Rightarrow \sin \theta +\sqrt{3}\cos \theta -\dfrac{1}{2}\cos \theta -\dfrac{\sqrt{3}}{2}\sin \theta =0 \\\ & \dfrac{1}{2}\left[ 2\sin \theta +2\sqrt{3}\cos \theta -\cos \theta -\sqrt{3}\sin \theta \right]=0 \\\ & \Rightarrow 2\sin \theta +2\sqrt{3}\cos \theta -\cos \theta -\sqrt{3}\sin \theta =0 \\\ \end{aligned}$$ Now let us make it in pairs of sine and cosine function. $$\begin{aligned} & 2\sin \theta -\sqrt{3}\sin \theta +2\sqrt{3}\cos \theta -\cos \theta =0 \\\ & \Rightarrow \left( 2-\sqrt{3} \right)\sin \theta +\left( 2\sqrt{3}-1 \right)\cos \theta =0 \\\ & \Rightarrow \left( 2-\sqrt{3} \right)\sin \theta =-\left( 2\sqrt{3}-1 \right)\cos \theta \\\ & \therefore \left( 2-\sqrt{3} \right)\sin \theta =\left( 1-2\sqrt{3} \right)\cos \theta \\\ \end{aligned}$$ We know that, $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$$. Thus from the above expression, $$\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-2\sqrt{3}}{2-\sqrt{3}}$$. $$\therefore $$ We can say that, $$\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-2\sqrt{3}}{2-\sqrt{3}}$$. i.e. $$\tan \theta =\dfrac{1-2\sqrt{3}}{2-\sqrt{3}}=0$$ - (3) From the question, it is said that, $$\tan \theta +\sqrt{3}=0$$. Thus put, $$\tan \theta =\dfrac{1-2\sqrt{3}}{2-\sqrt{3}}$$ in the above expression. $$\Rightarrow \dfrac{1-2\sqrt{3}}{2-\sqrt{3}}+\sqrt{3}=0$$ Let us simplify it, $$\begin{aligned} & \Rightarrow \dfrac{1-2\sqrt{3}+\left( 2-\sqrt{3} \right)\sqrt{3}}{2-\sqrt{3}}=\dfrac{1-2\sqrt{3}+2\sqrt{3}-\left( \sqrt{3}\times \sqrt{3} \right)}{2-\sqrt{3}} \\\ & \Rightarrow \dfrac{1-2\sqrt{3}+\left( 2-\sqrt{3} \right)\sqrt{3}}{2-\sqrt{3}}=\dfrac{1-\left( \sqrt{3}\times \sqrt{3} \right)}{2-\sqrt{3}}=\dfrac{1-3}{2-\sqrt{3}}=\dfrac{-2}{2-\sqrt{3}} \\\ \end{aligned}$$ Now let us rationalize it by multiplying the numerator and denominator by $$\left( 2+\sqrt{3} \right)$$. $$\therefore \dfrac{-2}{\left( 2-\sqrt{3} \right)}\times \dfrac{\left( 2+\sqrt{3} \right)}{\left( 2+\sqrt{3} \right)}=\dfrac{-2\left( 2+\sqrt{3} \right)}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}$$ ($$\because \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$$) $$\therefore \dfrac{-2}{\left( 2-\sqrt{3} \right)}\times \dfrac{\left( 2+\sqrt{3} \right)}{\left( 2+\sqrt{3} \right)}=\dfrac{-2\left( 2+\sqrt{3} \right)}{4-3}=-2\left( 2+\sqrt{3} \right)$$ From (3), we have, $$\begin{aligned} & \Rightarrow \tan \theta =\dfrac{1-2\sqrt{3}}{2-\sqrt{3}} \\\ & \therefore \theta ={{\tan }^{-1}}\left( \dfrac{1-2\sqrt{3}}{2-\sqrt{3}} \right) \\\ \end{aligned}$$ Hence we got the value of $$\theta ={{\tan }^{-1}}\left( \dfrac{1-2\sqrt{3}}{2-\sqrt{3}} \right)$$. Note: For solving a question like this, it is important that you know the basic trigonometric identities and trigonometric table by heart. Usually mistakes that students make is by forgetting the basic values and then guessing them. For example here they might guess that $$\cos \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$$. They can use technique that value of cos decreases with increase in angle. Trigonometry is one of the main topics and if you know the formula’s it’s easier to solve.