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Question: How do you verify the identity \(\dfrac{{\cos t + \cos 3t}}{{\sin 3t - \sin t}} = \cot t\)?...

How do you verify the identity cos⁡t+cos⁡3tsin⁡3t−sin⁡t=cot⁡t\dfrac{{\cos t + \cos 3t}}{{\sin 3t - \sin t}} = \cot t?

Explanation

Solution

To solve this problem, we will use the sum of products formula of trigonometry. This formula gives us the following forms if we add or subtract two cosine or sine functions. For this problem we will need the following two forms of the formula, that gives,
cos⁡a+cos⁡b=12cos⁡(a+b2)cos⁡(a−b2)\cos a + \cos b = \dfrac{1}{2}\cos \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)
⇒sin⁡a−sin⁡b=12sin⁡(a−b2)cos⁡(a+b2)\Rightarrow \sin a - \sin b = \dfrac{1}{2}\sin \left( {\dfrac{{a - b}}{2}} \right)\cos \left( {\dfrac{{a + b}}{2}} \right)
So, we will use these formulas on the left hand side of the identity and will try to derive the form on the right hand side.

Complete step by step answer:
We are given the identity, cos⁡t+cos⁡3tsin⁡3t−sin⁡t=cot⁡t\dfrac{{\cos t + \cos 3t}}{{\sin 3t - \sin t}} = \cot t. Now, operating the left hand side, we get,
⇒cos⁡t+cos⁡3tsin⁡3t−sin⁡t\Rightarrow \dfrac{{\cos t + \cos 3t}}{{\sin 3t - \sin t}}
⇒cos⁡3t+cos⁡tsin⁡3t−sin⁡t\Rightarrow \dfrac{{\cos 3t + \cos t}}{{\sin 3t - \sin t}}
We know, the sum of product formulas, that gives,
cos⁡a+cos⁡b=12cos⁡(a+b2)cos⁡(a−b2)\cos a + \cos b = \dfrac{1}{2}\cos \left( {\dfrac{{a + b}}{2}} \right)\cos \left( {\dfrac{{a - b}}{2}} \right)
⇒sin⁡a−sin⁡b=12sin⁡(a−b2)cos⁡(a+b2)\Rightarrow \sin a - \sin b = \dfrac{1}{2}\sin \left( {\dfrac{{a - b}}{2}} \right)\cos \left( {\dfrac{{a + b}}{2}} \right)

Therefore, using these formulas on the above function, we get,
⇒12cos⁡(3t+t2)cos⁡(3t−t2)12sin⁡(3t−t2)cos⁡(3t+t2)\Rightarrow \dfrac{{\dfrac{1}{2}\cos \left( {\dfrac{{3t + t}}{2}} \right)\cos \left( {\dfrac{{3t - t}}{2}} \right)}}{{\dfrac{1}{2}\sin \left( {\dfrac{{3t - t}}{2}} \right)\cos \left( {\dfrac{{3t + t}}{2}} \right)}}
Now, cancelling the common terms from the numerator and denominator, we get,
⇒cos⁡(3t+t2)cos⁡(3t−t2)sin⁡(3t−t2)cos⁡(3t+t2)\Rightarrow \dfrac{{\cos \left( {\dfrac{{3t + t}}{2}} \right)\cos \left( {\dfrac{{3t - t}}{2}} \right)}}{{\sin \left( {\dfrac{{3t - t}}{2}} \right)\cos \left( {\dfrac{{3t + t}}{2}} \right)}}

Now, simplifying the terms, we get,
⇒cos⁡(4t2)cos⁡(2t2)sin⁡(2t2)cos⁡(4t2)\Rightarrow \dfrac{{\cos \left( {\dfrac{{4t}}{2}} \right)\cos \left( {\dfrac{{2t}}{2}} \right)}}{{\sin \left( {\dfrac{{2t}}{2}} \right)\cos \left( {\dfrac{{4t}}{2}} \right)}}
⇒cos⁡(2t)cos⁡(t)sin⁡(t)cos⁡(2t)\Rightarrow \dfrac{{\cos \left( {2t} \right)\cos \left( t \right)}}{{\sin \left( t \right)\cos \left( {2t} \right)}}
Again, cancelling the common terms from the numerator and denominator, we get,
⇒cos⁡(t)sin⁡(t)\Rightarrow \dfrac{{\cos \left( t \right)}}{{\sin \left( t \right)}}
We know, cos⁡xsin⁡x=cot⁡x\dfrac{{\cos x}}{{\sin x}} = \cot x.

Therefore, using this property, we get,
⇒cot⁡t\Rightarrow \cot t
Now, the right hand side is given,
cot⁡t\cot t
Therefore, we can easily conclude that,
LHS = RHS
∴cos⁡t+cos⁡3tsin⁡3t−sin⁡t=cot⁡t\therefore \dfrac{{\cos t + \cos 3t}}{{\sin 3t - \sin t}} = \cot t
Hence, proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae and identities such as the sum of product formulas for trigonometric functions. We also need knowledge of algebraic rules and identities to simplify the expression. We must multiply the rational function with the same number in numerator and denominator to get to the required equality.