Question

Prove the following:
\[\sin \left( -{{420}^{o}} \right)\cos {{390}^{o}}+\cos \left( -{{660}^{o}} \right)\sin {{330}^{o}}=-1\]

Answer 1

Hint: First of all, take the LHS of the given equation. Now use \[\sin \left( 2n\pi +\theta \right)=\sin \theta ,\cos \left( 2n\pi \pm \theta \right)=\cos \theta \] and \[\sin \left( 2n\pi -\theta \right)=-\sin \theta \] and then take the required value of n and \[\theta \] to reduce the given expression. Now get the values of \[\sin {{30}^{o}},\sin {{60}^{o}},etc.\] from the trigonometric table to prove the desired result.

Complete step-by-step answer:

In this question, we have to prove that \[\sin \left( -{{420}^{o}} \right)\cos {{390}^{o}}+\cos

\left( -{{660}^{o}} \right)\sin {{330}^{o}}=-1\]. Let us consider the LHS of the equation given in the question.

\[LHS=\sin \left( -{{420}^{o}} \right)\cos {{390}^{o}}+\cos \left( -{{660}^{o}} \right)\sin

{{330}^{o}}\]

We know that, \[\sin \left( -\theta \right)=-\sin \theta \text{ and }\cos \left( -\theta

\right)=\cos \theta \]. By using these in the above equation, we get,

\[LHS=-\sin \left( {{420}^{o}} \right)\cos {{390}^{o}}+\cos \left( {{660}^{o}} \right)\sin

{{330}^{o}}\]

\[LHS=-\sin \left( {{360}^{o}}+{{60}^{o}} \right)\cos \left( {{360}^{o}}+{{30}^{o}} \right)+\cos

 \left( {{720}^{o}}-{{60}^{o}} \right)\sin \left( {{360}^{o}}-{{30}^{o}} \right).....\left( i \right)\]

We know that \[\sin \left( 2n\pi +\theta \right)=\sin \theta \]. So, by substituting \[n=1,\pi

={{180}^{o}}\] and \[\theta ={{60}^{o}}\], we get,

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

Also, we know that \[\cos \left( 2n\pi +\theta \right)=\cos \theta \]. So, by substituting

\[n=1,\pi ={{180}^{o}}\] and \[\theta ={{30}^{o}}\], we get,

\[\cos \left( {{360}^{o}}+{{30}^{o}} \right)=\cos {{30}^{o}}\]

By substituting the values of \[\cos \left( {{390}^{o}} \right)\] and \[\sin \left( {{420}^{o}} \right)\] in equation (i), we get,

\[LHS=\left( -\sin {{60}^{o}} \right)\left( \cos {{30}^{o}} \right)+\left( \cos{{720}^{o}}-{{60}^{o}} \right)\left( \sin {{360}^{o}}-{{30}^{o}} \right)....\left( ii \right)\]

We know that, \[\sin \left( 2n\pi -\theta \right)=-\sin \theta \]. So, by substituting \[n=1,\pi = {{180}^{o}}\] and \[\theta ={{30}^{o}}\], we get,

\[\sin \left( {{360}^{o}}-{{30}^{o}} \right)=-\sin {{30}^{o}}\]

Also, we know that \[\cos \left( 2n\pi -\theta \right)=\cos \theta \]. So, by substituting

\[n=2,\pi ={{180}^{o}}\] and \[\theta ={{60}^{o}}\], we get,

\[\cos \left( {{720}^{o}}-{{60}^{o}} \right)=\cos {{60}^{o}}\]

By substituting the values of \[\cos \left( {{660}^{o}} \right)\] and \[\sin \left( {{330}^{o}} \right)\] in equation (ii), we get,

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

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

Now, let us find the values of \[\sin {{60}^{o}},\sin {{30}^{o}},\cos {{30}^{o}}\] 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
\[\sin {{60}^{o}}=\dfrac{\sqrt{3}}{2}\]
\[\cos {{60}^{o}}=\dfrac{1}{2}\]
\[\sin {{30}^{o}}=\dfrac{1}{2}\]
\[\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}\]

By substituting these values in equation (iii), we get,
\[LHS=-\left[ \dfrac{\sqrt{3}}{2}.\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}+\dfrac{1}{2} \right]\]
\[LHS=-\left[ \dfrac{3}{4}+\dfrac{1}{4} \right]\]
\[LHS=-\dfrac{4}{4}\]
\[LHS=-1=RHS\]
So, we get, LHS = RHS

Hence proved.

Therefore, we have proved that

\[\sin \left( -{{420}^{o}} \right)\cos {{390}^{o}}+\cos \left( -{{660}^{o}} \right)\sin {{330}^{o}}=-1\]

Note: Students are advised to remember the general formulas like \[\sin \left( 2n\pi +\theta \right)=\sin \theta ,\cos \left( 2n\pi \pm \theta \right)=\cos \theta \], etc. Also, students should always convert the higher angles into their principal values that is between 0 to \[2\pi \] and then between 0 to \[\dfrac{\pi }{2}\] to easily solve the question because we only know the values of the trigonometric ratios at these angles.