Question

Prove the following
\[\cos {{510}^{\circ }}\cos {{330}^{\circ }}+\sin {{390}^{\circ }}\cos {{120}^{\circ }}=-1\]

Answer 1

Hint: To solve this question, we will consider the LHS of the above equation and applying the following trigonometric formula, we will get the RHS of the equation \[\cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta ,\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta ,\cos \left( {{360}^{\circ }}-\theta \right)=+\cos \theta ,\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta .\]

Complete step by step answer:
We are given that
\[\cos {{510}^{\circ }}\cos {{330}^{\circ }}+\sin {{390}^{\circ }}\cos {{120}^{\circ }}=-1.....\left( i \right)\]
To solve this question, let us consider the LHS of the above equation and make it equal to the RHS of the given equation (i). The LHs of the equation (i) is
\[\cos {{510}^{\circ }}\cos {{330}^{\circ }}+\sin {{390}^{\circ }}\cos {{120}^{\circ }}\]
Consider, \[\cos {{510}^{\circ }}=\cos \left( {{360}^{\circ }}+{{150}^{\circ }} \right)\left[ \text{As }{{510}^{\circ }}={{360}^{\circ }}+{{150}^{\circ }} \right]\]
\[\Rightarrow \cos {{510}^{\circ }}=\cos \left( {{360}^{\circ }}+{{150}^{\circ }} \right)\]
Now using the trigonometric identity, \[\cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta \] in the above equation using \[\theta ={{150}^{\circ }},\] we have,
\[\Rightarrow \cos {{510}^{\circ }}=\cos \left( {{360}^{\circ }}+{{150}^{\circ }} \right)\]
\[\Rightarrow \cos {{510}^{\circ }}=\cos {{150}^{\circ }}\]
Again splitting \[{{150}^{\circ }}\] as \[{{180}^{\circ }}-{{30}^{\circ }}\] we have
\[\Rightarrow \cos {{150}^{\circ }}=\cos \left( {{180}^{\circ }}-{{30}^{\circ }} \right)\]
Using the trigonometric identity, \[\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \] in the above obtained equation using \[\theta ={{30}^{\circ }},\] we have,
\[\cos \left( {{150}^{\circ }} \right)=\cos \left( {{180}^{\circ }}-{{30}^{\circ }} \right)=-\cos {{30}^{\circ }}\]
Similarly, we will solve \[\cos {{330}^{\circ }}\] and simplify it using trigonometric identities.
\[\Rightarrow \cos {{330}^{\circ }}=\cos \left( {{360}^{\circ }}-{{30}^{\circ }} \right)\]
As \[{{330}^{\circ }}={{360}^{\circ }}-{{30}^{\circ }}.\]
Using the identity, \[\cos \left( {{360}^{\circ }}-\theta \right)=\cos \theta \] in the above we have,
\[\Rightarrow \cos {{330}^{\circ }}=\cos \left( {{360}^{\circ }}-{{30}^{\circ }} \right)\]
\[\Rightarrow \cos {{330}^{\circ }}=\cos {{30}^{\circ }}\]
Hence, we have,
\[\cos {{510}^{\circ }}\cos {{330}^{\circ }}=-\cos {{30}^{\circ }}\cos {{30}^{\circ }}\]
\[\Rightarrow \cos {{510}^{\circ }}\cos {{330}^{\circ }}=-{{\cos }^{2}}{{30}^{\circ }}\]
Now consider the second part of the equation (i). We have,
\[\sin {{390}^{\circ }}=\sin \left( {{360}^{\circ }}+{{30}^{\circ }} \right)\]
\[\text{As }{{390}^{\circ }}={{360}^{\circ }}+{{30}^{\circ }}\]
Using the trigonometric identity, \[\sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta \] and taking \[\theta ={{30}^{\circ }},\] we have,
\[\sin {{390}^{\circ }}=\sin \left( {{360}^{\circ }}+{{30}^{\circ }} \right)=\sin {{30}^{\circ }}\]
Similarly, consider \[\cos {{120}^{\circ }}.\]
\[\cos {{120}^{\circ }}=\cos \left( {{90}^{\circ }}+{{30}^{\circ }} \right)\]
Using the trigonometric formula, \[\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta .\] We have,
\[\cos {{120}^{\circ }}=\cos \left( {{90}^{\circ }}+{{30}^{\circ }} \right)=-\sin {{30}^{\circ }}\]
\[\Rightarrow LHS=\cos {{510}^{\circ }}.\cos {{330}^{\circ }}+\sin {{390}^{\circ }}\cos {{120}^{\circ }}\]
\[\Rightarrow LHS=-\left( \cos {{30}^{\circ }} \right)\left( \cos {{30}^{\circ }} \right)+\sin {{30}^{\circ }}\left( -\sin {{30}^{\circ }} \right)\]
\[\Rightarrow LHS=-{{\cos }^{2}}{{30}^{\circ }}+\left( -{{\sin }^{2}}{{30}^{\circ }} \right)\]
\[\Rightarrow LHS=-\left( {{\cos }^{2}}{{30}^{\circ }}+{{\sin }^{2}}{{30}^{\circ }} \right)\]
\[\Rightarrow LHS=-1\]
As, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\forall \theta .\]
Hence, we have proved, \[\cos {{510}^{\circ }}\cos {{330}^{\circ }}+\sin {{390}^{\circ }}\cos {{120}^{\circ }}=-1.\]

Note: At the end where we have obtained \[-{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{30}^{\circ }}\] we can also substitute the value of \[\sin {{30}^{\circ }}\] and \[\cos {{30}^{\circ }}\] and without using \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] to get the result.
\[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2};\sin {{30}^{\circ }}=\dfrac{1}{2}\]
\[\Rightarrow {{\cos }^{2}}{{30}^{\circ }}=\dfrac{3}{4};{{\sin }^{2}}{{30}^{\circ }}=\dfrac{1}{4}\]
\[\Rightarrow {{\cos }^{2}}{{30}^{\circ }}+{{\sin }^{2}}{{30}^{\circ }}=\dfrac{3}{4}+\dfrac{1}{4}\]
\[\Rightarrow {{\cos }^{2}}{{30}^{\circ }}+{{\sin }^{2}}{{30}^{\circ }}=\dfrac{4}{4}\]
\[\Rightarrow {{\cos }^{2}}{{30}^{\circ }}+{{\sin }^{2}}{{30}^{\circ }}=1\]
And hence, \[-\left( {{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }} \right)=-1.\] Hence proved.