Solve the given trigonometric equation: ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+{{120}^{\circ }} \right)+{{\cos }^{2}}\left( A-{{120}^{\circ }} \right)=\dfrac{3}{2}$
Answer
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Hint: Simplify the given trigonometric expression using the trigonometric identity $\cos x=-\cos \left( {{180}^{\circ }}-x \right)$ and $\sin \left( {{90}^{\circ }}-x \right)=\cos x$. Use another trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to further simplify the given expression and thus calculate the value of A which satisfies the equation.
Complete step-by-step answer:
We have to find the value of A which satisfies the given trigonometric equation ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+{{120}^{\circ }} \right)+{{\cos }^{2}}\left( A-{{120}^{\circ }} \right)=\dfrac{3}{2}$.
We know the trigonometric identity $\cos x=-\cos \left( {{180}^{\circ }}-x \right)$.
Substituting $x={{60}^{\circ }}-A$ in the above equation, we have $\cos \left( {{60}^{\circ }}-A \right)=-\cos \left( {{180}^{\circ }}-\left( {{60}^{\circ }}-A \right) \right)=-\cos \left( A+{{120}^{\circ }} \right)$.
Squaring the above equation on both sides, we have ${{\cos }^{2}}\left( {{60}^{\circ }}-A \right)={{\cos }^{2}}\left( A+{{120}^{\circ }} \right).....\left( 1 \right)$.
We know the trigonometric identity $\cos x=-\cos \left( {{180}^{\circ }}-x \right)$.
Substituting $x={{30}^{\circ }}+A$ in the above equation, we have $\cos \left( {{90}^{\circ }}+{{30}^{\circ }}+A \right)=\cos \left( {{120}^{\circ }}+A \right)=-\sin \left( {{30}^{\circ }}+A \right)$.
Squaring the above equation on both sides, we have ${{\cos }^{2}}\left( {{120}^{\circ }}+A \right)={{\sin }^{2}}\left( {{30}^{\circ }}+A \right).....\left( 2 \right)$.
Substituting equation (1) and (2) in the given trigonometric equation, we have ${{\cos }^{2}}A+{{\cos }^{2}}\left( {{60}^{\circ }}-A \right)+{{\sin }^{2}}\left( {{30}^{\circ }}+A \right)=\dfrac{3}{2}.....\left( 3 \right)$.
We know the trigonometric identity $\sin \left( {{90}^{\circ }}-x \right)=\cos x$.
Substituting $x={{60}^{\circ }}-A$ in the above equation, we have $\sin \left( {{90}^{\circ }}-\left( {{60}^{\circ }}-A \right) \right)=\sin \left( A+{{30}^{\circ }} \right)=\cos \left( {{60}^{\circ }}-A \right)$.
Squaring the above equation on both sides, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\cos }^{2}}\left( {{60}^{\circ }}-A \right).....\left( 4 \right)$.
Substituting equation (4) in equation (3), we have ${{\cos }^{2}}A+2{{\sin }^{2}}\left( {{30}^{\circ }}+A \right)=\dfrac{3}{2}.....\left( 5 \right)$.
We know the trigonometric identity $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$.
Substituting $x={{30}^{\circ }},y=A$ in the above equation, we have $\sin \left( {{30}^{\circ }}+A \right)=\sin {{30}^{\circ }}\cos A+\cos {{30}^{\circ }}\sin A=\dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2}$.
Squaring the above equation on both sides, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}$.
We know the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Substituting $a=\dfrac{\cos A}{2},b=\dfrac{\sqrt{3}\sin A}{2}$ in the above equation, we have ${{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}=\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2}$.
Thus, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}=\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2}.....\left( 6 \right)$.
Substituting equation (6) in equation (5), we have ${{\cos }^{2}}A+2\left( \dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2} \right)=\dfrac{3}{2}$.
Simplifying the above equation, we have ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3{{\sin }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$.
Thus, we can rewrite the equation ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3{{\sin }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$ as ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3\left( 1-{{\cos }^{2}}A \right)}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
Simplifying the above equation, we have $\dfrac{3{{\cos }^{2}}A}{2}+\dfrac{3}{2}-\dfrac{3{{\cos }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
Thus, we have $\dfrac{3}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}\Rightarrow \sqrt{3}\sin A\cos A=0$.
Multiplying the equation $\sqrt{3}\sin A\cos A=0$ by 2 on both sides, we have $2\sqrt{3}\sin A\cos A=0$.
We know the trigonometric identity $\sin 2x=2\sin x\cos x$.
Thus, we can rewrite the equation $2\sqrt{3}\sin A\cos A=0$ as $\sqrt{3}\sin 2A=0\Rightarrow \sin 2A=0$.
We know that the solution of $\sin x=0$ is $x=n\pi ,n\in I$.
So, the solution of $\sin 2A=0$ is $2A=n\pi ,n\in I\Rightarrow A=\dfrac{n\pi }{2},n\in I$.
Hence, the value of ‘A’ which satisfies the given trigonometric equation is $A=\dfrac{n\pi }{2},n\in I$.
Note: One must keep in mind that we have to find out all the values of ‘A’ which satisfy the given equation. We can also solve this question by using the identity $\cos 2x=1-2{{\sin }^{2}}x$ and then simplifying the expression to calculate the value of ‘A’.
Complete step-by-step answer:
We have to find the value of A which satisfies the given trigonometric equation ${{\cos }^{2}}A+{{\cos }^{2}}\left( A+{{120}^{\circ }} \right)+{{\cos }^{2}}\left( A-{{120}^{\circ }} \right)=\dfrac{3}{2}$.
We know the trigonometric identity $\cos x=-\cos \left( {{180}^{\circ }}-x \right)$.
Substituting $x={{60}^{\circ }}-A$ in the above equation, we have $\cos \left( {{60}^{\circ }}-A \right)=-\cos \left( {{180}^{\circ }}-\left( {{60}^{\circ }}-A \right) \right)=-\cos \left( A+{{120}^{\circ }} \right)$.
Squaring the above equation on both sides, we have ${{\cos }^{2}}\left( {{60}^{\circ }}-A \right)={{\cos }^{2}}\left( A+{{120}^{\circ }} \right).....\left( 1 \right)$.
We know the trigonometric identity $\cos x=-\cos \left( {{180}^{\circ }}-x \right)$.
Substituting $x={{30}^{\circ }}+A$ in the above equation, we have $\cos \left( {{90}^{\circ }}+{{30}^{\circ }}+A \right)=\cos \left( {{120}^{\circ }}+A \right)=-\sin \left( {{30}^{\circ }}+A \right)$.
Squaring the above equation on both sides, we have ${{\cos }^{2}}\left( {{120}^{\circ }}+A \right)={{\sin }^{2}}\left( {{30}^{\circ }}+A \right).....\left( 2 \right)$.
Substituting equation (1) and (2) in the given trigonometric equation, we have ${{\cos }^{2}}A+{{\cos }^{2}}\left( {{60}^{\circ }}-A \right)+{{\sin }^{2}}\left( {{30}^{\circ }}+A \right)=\dfrac{3}{2}.....\left( 3 \right)$.
We know the trigonometric identity $\sin \left( {{90}^{\circ }}-x \right)=\cos x$.
Substituting $x={{60}^{\circ }}-A$ in the above equation, we have $\sin \left( {{90}^{\circ }}-\left( {{60}^{\circ }}-A \right) \right)=\sin \left( A+{{30}^{\circ }} \right)=\cos \left( {{60}^{\circ }}-A \right)$.
Squaring the above equation on both sides, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\cos }^{2}}\left( {{60}^{\circ }}-A \right).....\left( 4 \right)$.
Substituting equation (4) in equation (3), we have ${{\cos }^{2}}A+2{{\sin }^{2}}\left( {{30}^{\circ }}+A \right)=\dfrac{3}{2}.....\left( 5 \right)$.
We know the trigonometric identity $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$.
Substituting $x={{30}^{\circ }},y=A$ in the above equation, we have $\sin \left( {{30}^{\circ }}+A \right)=\sin {{30}^{\circ }}\cos A+\cos {{30}^{\circ }}\sin A=\dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2}$.
Squaring the above equation on both sides, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}$.
We know the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Substituting $a=\dfrac{\cos A}{2},b=\dfrac{\sqrt{3}\sin A}{2}$ in the above equation, we have ${{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}=\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2}$.
Thus, we have ${{\sin }^{2}}\left( {{30}^{\circ }}+A \right)={{\left( \dfrac{\cos A}{2}+\dfrac{\sqrt{3}\sin A}{2} \right)}^{2}}=\dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2}.....\left( 6 \right)$.
Substituting equation (6) in equation (5), we have ${{\cos }^{2}}A+2\left( \dfrac{{{\cos }^{2}}A}{4}+\dfrac{3{{\sin }^{2}}A}{4}+\dfrac{\sqrt{3}\sin A\cos A}{2} \right)=\dfrac{3}{2}$.
Simplifying the above equation, we have ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3{{\sin }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$.
Thus, we can rewrite the equation ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3{{\sin }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$ as ${{\cos }^{2}}A+\dfrac{{{\cos }^{2}}A}{2}+\dfrac{3\left( 1-{{\cos }^{2}}A \right)}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
Simplifying the above equation, we have $\dfrac{3{{\cos }^{2}}A}{2}+\dfrac{3}{2}-\dfrac{3{{\cos }^{2}}A}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}$.
Thus, we have $\dfrac{3}{2}+\sqrt{3}\sin A\cos A=\dfrac{3}{2}\Rightarrow \sqrt{3}\sin A\cos A=0$.
Multiplying the equation $\sqrt{3}\sin A\cos A=0$ by 2 on both sides, we have $2\sqrt{3}\sin A\cos A=0$.
We know the trigonometric identity $\sin 2x=2\sin x\cos x$.
Thus, we can rewrite the equation $2\sqrt{3}\sin A\cos A=0$ as $\sqrt{3}\sin 2A=0\Rightarrow \sin 2A=0$.
We know that the solution of $\sin x=0$ is $x=n\pi ,n\in I$.
So, the solution of $\sin 2A=0$ is $2A=n\pi ,n\in I\Rightarrow A=\dfrac{n\pi }{2},n\in I$.
Hence, the value of ‘A’ which satisfies the given trigonometric equation is $A=\dfrac{n\pi }{2},n\in I$.
Note: One must keep in mind that we have to find out all the values of ‘A’ which satisfy the given equation. We can also solve this question by using the identity $\cos 2x=1-2{{\sin }^{2}}x$ and then simplifying the expression to calculate the value of ‘A’.
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