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Question: If \({T_n} = {\sin ^n}x + {\cos ^n}x\) , prove that \(\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_...

If Tn=sin⁡nx+cos⁡nx{T_n} = {\sin ^n}x + {\cos ^n}x , prove that T3−T5T1=T5−T7T3\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}} .

Explanation

Solution

It is given in the question that Tn=sin⁡nx+cos⁡nx{T_n} = {\sin ^n}x + {\cos ^n}x .
Then, we have to prove T3−T5T1=T5−T7T3\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}} .
First, we will find the equation of T1,T3,T5,T7{T_1},{T_3},{T_5},{T_7} by using Tn=sin⁡nx+cos⁡nx{T_n} = {\sin ^n}x + {\cos ^n}x .
Then after, put the values according to the given question.
Finally, solving further we will prove that T3−T5T1=T5−T7T3\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}} .

Complete step-by-step answer:
It is given in the question that Tn=sin⁡nx+cos⁡nx{T_n} = {\sin ^n}x + {\cos ^n}x .
Then, we have to prove T3−T5T1=T5−T7T3\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}} .
Since, Tn=sin⁡nx+cos⁡nx{T_n} = {\sin ^n}x + {\cos ^n}x
Now, according to above equation,
T1=sin⁡x+cos⁡x{T_1} = \sin x + \cos x
T3=sin⁡3x+cos⁡3x{T_3} = {\sin ^3}x + {\cos ^3}x
T5=sin⁡5x+cos⁡5x{T_5} = {\sin ^5}x + {\cos ^5}x
T7=sin⁡7x+cos⁡7x{T_7} = {\sin ^7}x + {\cos ^7}x
Now, according to question,
T3−T5=sin⁡3x+cos⁡3x−sin⁡5x−cos⁡5x{T_3} - {T_5} = {\sin ^3}x + {\cos ^3}x - {\sin ^5}x - {\cos ^5}x
T3−T5=sin⁡3x−sin⁡5x+cos⁡3x−cos⁡5x{T_3} - {T_5} = {\sin ^3}x - {\sin ^5}x + {\cos ^3}x - {\cos ^5}x
Now, take out sin⁡3x{\sin ^3}x and cos⁡3x{\cos ^3}x common, we get,
T3−T5=sin⁡3x(1−sin⁡2x)+cos⁡3x(1−cos⁡2x){T_3} - {T_5} = {\sin ^3}x\left( {1 - {{\sin }^2}x} \right) + {\cos ^3}x\left( {1 - {{\cos }^2}x} \right)
Since, we know that 1−sin⁡2x=cos⁡2x1 - {\sin ^2}x = {\cos ^2}x and 1−cos⁡2x=sin⁡2x1 - {\cos ^2}x = {\sin ^2}x .
T3−T5=sin⁡3x.cos⁡2x+cos⁡3x.sin⁡2x{T_3} - {T_5} = {\sin ^3}x.{\cos ^2}x + {\cos ^3}x.{\sin ^2}x
T3−T5=sin⁡2x.cos⁡2x(sin⁡x+cos⁡x){T_3} - {T_5} = {\sin ^2}x.{\cos ^2}x\left( {\sin x + \cos x} \right)
T3−T5T1=sin⁡2x.cos⁡2x(sin⁡x+cos⁡x)(sin⁡x+cos⁡x)\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{{\sin }^2}x.{{\cos }^2}x\left( {\sin x + \cos x} \right)}}{{\left( {\sin x + \cos x} \right)}}
∴T3−T5T1=sin⁡2x.cos⁡2x\therefore \dfrac{{{T_3} - {T_5}}}{{{T_1}}} = {\sin ^2}x.{\cos ^2}x
Now, similarly we will find the value of T5−T7T3\dfrac{{{T_5} - {T_7}}}{{{T_3}}}
T5−T7=sin⁡5x+cos⁡5x−sin⁡7x−cos⁡7x{T_5} - {T_7} = {\sin ^5}x + {\cos ^5}x - {\sin ^7}x - {\cos ^7}x
T5−T7=sin⁡5x−sin⁡7x+cos⁡5x−cos⁡7x{T_5} - {T_7} = {\sin ^5}x - {\sin ^7}x + {\cos ^5}x - {\cos ^7}x
Now, take out sin⁡5x{\sin ^5}x and cos⁡5x{\cos ^5}x common, we get,
T5−T7=sin⁡5x(1−sin⁡2x)+cos⁡5x(1−cos⁡2x){T_5} - {T_7} = {\sin ^5}x\left( {1 - {{\sin }^2}x} \right) + {\cos ^5}x\left( {1 - {{\cos }^2}x} \right)
Since, we know that 1−sin⁡2x=cos⁡2x1 - {\sin ^2}x = {\cos ^2}x and 1−cos⁡2x=sin⁡2x1 - {\cos ^2}x = {\sin ^2}x .
T5−T7=sin⁡5x.cos⁡2x+cos⁡5x.sin⁡2x{T_5} - {T_7} = {\sin ^5}x.{\cos ^2}x + {\cos ^5}x.{\sin ^2}x
T5−T7=sin⁡2x.cos⁡2x(sin⁡3x+cos⁡3x){T_5} - {T_7} = {\sin ^2}x.{\cos ^2}x\left( {{{\sin }^3}x + {{\cos }^3}x} \right)
T5−T7T3=sin⁡2x.cos⁡2x(sin⁡3x+cos⁡3x)(sin⁡3x+cos⁡3x)\dfrac{{{T_5} - {T_7}}}{{{T_3}}} = \dfrac{{{{\sin }^2}x.{{\cos }^2}x\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}}{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}}
T5−T7T3=sin⁡2x.cos⁡2x\dfrac{{{T_5} - {T_7}}}{{{T_3}}} = {\sin ^2}x.{\cos ^2}x .
Hence, T3−T5T1=T5−T7T3\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}} .

Note: There are various distinct trigonometric identities. When a trigonometric function is involved in an equation then trigonometric identities are useful to solve that equation.
We can use identities cos⁡ec2x−cot⁡2x=1\cos e{c^2}x - {\cot ^2}x = 1 and sec2x−tan⁡2x=1se{c^2}x - {\tan ^2}x = 1 to solve many trigonometric problems. These identities are called Pythagorean identities.