Question

Prove the trigonometric equation:
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$

Answer 1

Hint:In this case, we need to use the formulas for the addition of cosine and sine of two angles to simplify the numerator and the denominator. Thereafter, we can factor out the numerator and denominator to obtain the required expression in terms of $cos\left( 3x \right)$ and $sin\left( 3x \right)$. After that, we can use the definition of $cot$ to write the expression in terms of cot and obtain the required answer.

Complete step-by-step answer:
We notice that in the right hand side of the equation only the $cot$ of 3x appears whereas in the left hand side, 2x, 3x, 4x appear. Therefore, we should try to rewrite the Left Hand side so that we can write everything in terms of 3x. For this we can use the formula for addition of sines and cosines of two angles which is stated as
$\sin (a)+\sin (b)=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$
and
$\cos (a)+\cos (b)=2\cos \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$

Now, taking a=4x and b=2x in equations, and using it in the equation we obtain
$\begin{aligned} & \sin (4x)+\sin (2x)=2\sin \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right) \\\ & =2\sin 3x\cos x.................(1.1) \\\ \end{aligned}$
and
$\begin{aligned} & \cos (4x)+\cos (2x)=2\cos \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right) \\\ & =2\cos 3x\cos x.................................(1.2) \\\ \end{aligned}$
Therefore, using equations (1.1) and (1.2) the Left Hand Side of the equation given in the question can be written as
$\begin{aligned} & \dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\dfrac{\cos \left( 3x \right)+2\cos \left( 3x \right)\cos \left( x \right)}{\sin (3x)+2\sin (3x)\cos (x)} \\\ & =\dfrac{\cos 3x(1+2\cos x)}{\sin 3x(1+2\cos x)} \\\ \end{aligned}$
Now, we notice that the factor of $1+2cos\left( x \right)$ will get cancelled in the numerator and denominator to give
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\dfrac{\cos 3x(1+2\cos x)}{\sin 3x(1+2\cos x)}=\dfrac{\cos 3x}{\sin 3x}................(1.3)$

Now, we look at the definition of cot (short form of cotangent) of an angle a which is stated as
$\cot (\theta )=\dfrac{\cos (\theta )}{\sin (\theta )}...............(1.4)$
Taking $\theta =3x$ in equation1.4 and using it in equation 1.3, we obtain
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\dfrac{\cos 3x}{\sin 3x}=\cot 3x$
Which is the same expression in the Right Hand Side (RHS). Thus, we have proved the equation.

Note: While using equations (1.1) and (1.2) to obtain the addition of $cos$ and $sin$, we should be careful to add the terms involving 2x and 4x, as only then we can get 3x in the result which is present in the right hand side, taking other terms would complicate the problem and it would be difficult to obtain cot3x from the resulting expression.