# Prove the following : \[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x\]

Last updated date: 29th Mar 2023

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Answer

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Hint: Take the LHS of the expression. Apply the basic trigonometric identities in the numerator and denominator of the expression and simplify it. Hence, prove that LHS=RHS.

“Complete step-by-step answer:”

We have been given the expression, \[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x\].

Let us consider the LHS of the expression.

LHS = \[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}\].

We know the basic trigonometric identities,

\[\begin{align}

& \cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \\

& \sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \\

\end{align}\]

Let us take \[\left( \cos 4x+\cos 2x \right)\] from the numerator and apply the trigonometric identity.

\[\therefore \cos 4x+\cos 2x=2\cos \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right)\]

\[\begin{align}

& =2\cos \left( \dfrac{6x}{2} \right)\cos \left( \dfrac{2x}{2} \right) \\

& =2\cos 3x\cos x \\

\end{align}\]

Now let us substitute the values of \[\left( \cos 4x+\cos 2x \right)\] and \[\left( \sin 4x+\sin 2x \right)\] in the LHS.

\[\begin{align}

& LHS=\dfrac{\left( \cos 4x+\cos 2x \right)+\cos 3x}{\left( \sin 4x+\sin 2x \right)+\sin 3x} \\

& LHS=\dfrac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x} \\

\end{align}\]

Take \[\left( \cos 3x \right)\]common from the numerator and \[\left( \sin 3x \right)\]from the denominator.

\[=\dfrac{\cos 3x\left[ 2\cos x+1 \right]}{\sin 3x\left[ 2\cos x+1 \right]}\]

Cancel out \[\left( 2\cos x+1 \right)\] from the numerator and denominator.

\[\therefore LHS=\dfrac{\cos 3x}{\sin 3x}\]

We know that, \[\dfrac{\cos x}{\sin x}=\cot x\]

Hence, \[\dfrac{\cos 3x}{\sin 3x}=\cot 3x\].

\[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x\]

\[\therefore \]LHS = RHS

Hence proved.

Note: Remember the basic trigonometric identities like \[\left( \sin a+\sin b \right)\] and \[\left( \cos a+\cos b \right)\] which we have used to solve this expression. Trigonometric identities are an important section in maths. Just apply the formula and simplify it to get the required answer.

“Complete step-by-step answer:”

We have been given the expression, \[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x\].

Let us consider the LHS of the expression.

LHS = \[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}\].

We know the basic trigonometric identities,

\[\begin{align}

& \cos a+\cos b=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \\

& \sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \\

\end{align}\]

Let us take \[\left( \cos 4x+\cos 2x \right)\] from the numerator and apply the trigonometric identity.

\[\therefore \cos 4x+\cos 2x=2\cos \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right)\]

\[\begin{align}

& =2\cos \left( \dfrac{6x}{2} \right)\cos \left( \dfrac{2x}{2} \right) \\

& =2\cos 3x\cos x \\

\end{align}\]

Now let us substitute the values of \[\left( \cos 4x+\cos 2x \right)\] and \[\left( \sin 4x+\sin 2x \right)\] in the LHS.

\[\begin{align}

& LHS=\dfrac{\left( \cos 4x+\cos 2x \right)+\cos 3x}{\left( \sin 4x+\sin 2x \right)+\sin 3x} \\

& LHS=\dfrac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x} \\

\end{align}\]

Take \[\left( \cos 3x \right)\]common from the numerator and \[\left( \sin 3x \right)\]from the denominator.

\[=\dfrac{\cos 3x\left[ 2\cos x+1 \right]}{\sin 3x\left[ 2\cos x+1 \right]}\]

Cancel out \[\left( 2\cos x+1 \right)\] from the numerator and denominator.

\[\therefore LHS=\dfrac{\cos 3x}{\sin 3x}\]

We know that, \[\dfrac{\cos x}{\sin x}=\cot x\]

Hence, \[\dfrac{\cos 3x}{\sin 3x}=\cot 3x\].

\[\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x\]

\[\therefore \]LHS = RHS

Hence proved.

Note: Remember the basic trigonometric identities like \[\left( \sin a+\sin b \right)\] and \[\left( \cos a+\cos b \right)\] which we have used to solve this expression. Trigonometric identities are an important section in maths. Just apply the formula and simplify it to get the required answer.

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