Question

Prove the following:
\[\sin {{780}^{o}}\sin {{480}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}=\dfrac{1}{2}\]

Answer 1

Hint: First of all, take the LHS of the given equation and convert the angles in between 0 to \[{{90}^{o}}\] by using \[\sin \left( 2n\pi +\theta \right)=\sin \theta ,\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \] and \[\cos \left( {{180}^{o}}-\theta \right)=-\cos \theta \] and then use a trigonometric table for the general angles to prove the desired result.

Complete step-by-step answer:

In this question, we have to prove that \[\sin {{780}^{o}}\sin {{480}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}=\dfrac{1}{2}\]. Let us consider the LHS of the equation given in the question.

\[LHS=\sin {{780}^{o}}\sin {{480}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}\]

We know that,

\[{{780}^{o}}=\left( 2\times {{360}^{o}} \right)+{{60}^{o}}\]

\[{{480}^{o}}={{360}^{o}}+{{120}^{o}}\]

By using these in the above equation, we get,

\[LHS=\sin \left( 2\times {{360}^{o}}+{{60}^{o}} \right)\sin \left( {{360}^{o}}+{{120}^{o}}

\right)+\cos {{120}^{o}}\sin {{150}^{o}}......\left( i \right)\]

We know that \[\sin \left( 2n\pi +\theta \right)=\sin \theta \] where \[\theta \in W\]. So, by

substituting \[n=1,\pi ={{180}^{o}}\] and \[\theta ={{60}^{o}}\], we get,

 \[\sin \left( 2\times {{360}^{o}}+{{60}^{o}} \right)=\sin {{60}^{o}}\]

Also, by substituting \[n=1,\pi ={{180}^{o}}\] and \[\theta ={{120}^{o}}\], we get,

\[\sin \left( {{360}^{o}}+{{120}^{o}} \right)=\sin {{120}^{o}}\]

By substituting these values in equation (i), we get,

\[LHS=\left( \sin {{60}^{o}} \right)\left( \sin {{120}^{o}} \right)+\left( \cos {{120}^{o}}

\right)\left( \sin {{150}^{o}} \right)\]

We can also write the above equation as,

\[LHS=\left( \sin {{60}^{o}} \right)\left[ \sin \left( {{180}^{o}}-{{60}^{o}} \right) \right]+\left[ \cos \left( {{180}^{o}}-{{60}^{o}} \right) \right]\left[ \sin \left( {{180}^{o}}-{{30}^{o}} \right) \right]\]

We know that \[\sin \left( {{180}^{o}}-\theta \right)=\sin \theta \] and \[\cos \left(

{{180}^{o}}-\theta \right)=-\cos \theta \]. By using these in the above equation, we get,


\[LHS=\left( \sin {{60}^{o}} \right)\left( \sin {{60}^{o}} \right)+\left( -\cos {{60}^{o}}

\right)\left( \sin {{30}^{o}} \right)....\left( ii \right)\]

Now, let us find the values of \[\sin {{60}^{o}},\sin {{30}^{o}}\] and \[\cos {{60}^{o}}\] from the trigonometric table for general angles.

	\[\sin \theta \]	\[\cos \theta \]	\[\tan \theta \]	\[\operatorname{cosec}\theta \]	\[\sec \theta \]	\[\cot \theta \]
0	0	1	0	-	1	-
\[\dfrac{\pi }{6}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{3}}\]	2	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{3}\]
\[\dfrac{\pi }{4}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]	1	\[\sqrt{2}\]	\[\sqrt{2}\]	1
\[\dfrac{\pi }{3}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]	\[\sqrt{3}\]	\[\dfrac{2}{\sqrt{3}}\]	2	\[\dfrac{1}{\sqrt{3}}\]
\[\dfrac{\pi }{2}\]	1	0	-	1	-	0

From the above table, we get

\[\sin {{60}^{o}}=\dfrac{\sqrt{3}}{2}\]

\[\cos {{60}^{o}}=\dfrac{1}{2}\]

\[\sin {{30}^{o}}=\dfrac{1}{2}\]

By substituting these values in equation (ii), we get,

\[LHS=\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{-1}{2}
\right)\left( \dfrac{1}{2} \right)\]

\[LHS={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}\]

\[LHS=\dfrac{3}{4}-\dfrac{1}{4}\]

\[LHS=\dfrac{2}{4}\]

\[LHS=\dfrac{1}{2}=RHS\]

So, we get, LHS = RHS

Hence proved.

Therefore, we have proved that
\[\sin {{780}^{o}}\sin {{480}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}}=\dfrac{1}{2}\]

Note: In these types of questions, students should always try to convert the higher angles between 0 to \[{{90}^{o}}\] because, for these angles only, we know the value for different trigonometric ratios. Also, students often take \[\cos \left( 180-\theta \right)=\cos \theta \] which is wrong because \[\cos \left( 180-\theta \right)=-\cos \theta \] while \[\sin \left( 180-\theta \right)=\sin \theta \]. So, this must be taken care of. Students should also memorize the table for the general trigonometric ratios at least for \[\sin \theta ,\cos \theta \] and \[\tan \theta \].