Question

If ${T_n} = {\sin ^n}x + {\cos ^n}x$, prove that $\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}}$ .

Answer 1

Hint: In this problem, first we will find L.H.S. part $\dfrac{{{T_3} - {T_5}}}{{{T_1}}}$ by putting $n = 1,3,5$ in ${T_n} = {\sin ^n}x + {\cos ^n}x$. Then, we will find R.H.S. part $\dfrac{{{T_5} - {T_7}}}{{{T_3}}}$ by putting $n = 3,5,7$ in ${T_n} = {\sin ^n}x + {\cos ^n}x$. Also we will use the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

Complete step by step solution: In this problem, it is given that ${T_n} = {\sin ^n}x + {\cos ^n}x\; \cdots \cdots \left( 1 \right)$.
Let us find ${T_1}$ by putting $n = 1$ in equation $\left( 1 \right)$. Therefore, we get ${T_1} = {\sin ^1}x + {\cos ^1}x = \sin x + \cos x\; \cdots \cdots \left( 2 \right)$
Let us find ${T_3}$ by putting $n = 3$ in equation $\left( 1 \right)$. Therefore, we get ${T_3} = {\sin ^3}x + {\cos ^3}x\; \cdots \cdots \left( 3 \right)$
Let us find ${T_5}$ by putting $n = 5$ in equation $\left( 1 \right)$. Therefore, we get ${T_5} = {\sin ^5}x + {\cos ^5}x\; \cdots \cdots \left( 4 \right)$
Now we are going to find L.H.S. part $\dfrac{{{T_3} - {T_5}}}{{{T_1}}}$ by using equations $\left( 2 \right),\left( 3 \right)$ and $\left( 4 \right)$.
L.H.S. $ = \dfrac{{{T_3} - {T_5}}}{{{T_1}}}$
$ = \dfrac{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right) - \left( {{{\sin }^5}x + {{\cos }^5}x} \right)}}{{\sin x + \cos x}}$
$ = \dfrac{{{{\sin }^3}x + {{\cos }^3}x - {{\sin }^5}x - {{\cos }^5}x}}{{\sin x + \cos x}}$
Rewrite the above equation, we get
$ = \dfrac{{{{\sin }^3}x - {{\sin }^5}x + {{\cos }^3}x - {{\cos }^5}x}}{{\sin x + \cos x}}$
$ = \dfrac{{{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right) + {{\cos }^3}x\left( {1 - {{\cos }^2}x} \right)}}{{\sin x + \cos x}}$
Now we are going to use Pythagorean identity ${\sin ^2}x + {\cos ^2}x = 1$. Note that ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$ or ${\cos ^2}x = 1 - {\sin ^2}x$.
$ \Rightarrow $ L.H.S. $ = \dfrac{{{{\sin }^3}x\left( {{{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x} \right)}}{{\sin x + \cos x}}$
Taking ${\sin ^2}x \cdot {\cos ^2}x$ common out from the numerator, we get
L.H.S. $ = \dfrac{{{{\sin }^2}x \cdot {{\cos }^2}x\left( {\sin x + \cos x} \right)}}{{\sin x + \cos x}}$
On cancellation of the factor $\sin x + \cos x$, we get
L.H.S. $ = {\sin ^2}x \cdot {\cos ^2}x$
Therefore, we get $\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = {\sin ^2}x \cdot {\cos ^2}x$.
Let us find ${T_7}$ by putting $n = 7$ in equation $\left( 1 \right)$. Therefore, we get ${T_7} = {\sin ^7}x + {\cos ^7}x\; \cdots \cdots \left( 5 \right)$.
Now we are going to find R.H.S. part $\dfrac{{{T_5} - {T_7}}}{{{T_3}}}$ by using equations $\left( 3 \right),\left( 4 \right)$ and $\left( 5 \right)$.
R.H.S. $ = \dfrac{{{T_5} - {T_7}}}{{{T_3}}}$
$ = \dfrac{{\left( {{{\sin }^5}x + {{\cos }^5}x} \right) - \left( {{{\sin }^7}x + {{\cos }^7}x} \right)}}{{{{\sin }^3}x + {{\cos }^3}x}}$
$ = \dfrac{{{{\sin }^5}x + {{\cos }^5}x - {{\sin }^7}x - {{\cos }^7}x}}{{{{\sin }^3}x + {{\cos }^3}x}}$
Rewrite the above equation, we get
$ = \dfrac{{{{\sin }^5}x - {{\sin }^7}x + {{\cos }^5}x - {{\cos }^7}x}}{{{{\sin }^3}x + {{\cos }^3}x}}$
$ = \dfrac{{{{\sin }^5}x\left( {1 - {{\sin }^2}x} \right) + {{\cos }^5}x\left( {1 - {{\cos }^2}x} \right)}}{{{{\sin }^3}x + {{\cos }^3}x}}$
Now we are going to use Pythagorean identity ${\sin ^2}x + {\cos ^2}x = 1$. Note that ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$ or ${\cos ^2}x = 1 - {\sin ^2}x$.
$ \Rightarrow $ R.H.S. $ = \dfrac{{{{\sin }^5}x\left( {{{\cos }^2}x} \right) + {{\cos }^5}x\left( {{{\sin }^2}x} \right)}}{{{{\sin }^3}x + {{\cos }^3}x}}$
Taking ${\sin ^2}x \cdot {\cos ^2}x$ common out from the numerator, we get
R.H.S. $ = \dfrac{{{{\sin }^2}x \cdot {{\cos }^2}x\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}}{{{{\sin }^3}x + {{\cos }^3}x}}$
On cancellation of the factor ${\sin ^3}x + {\cos ^3}x$, we get
R.H.S. $ = {\sin ^2}x \cdot {\cos ^2}x$
Therefore, we get $\dfrac{{{T_5} - {T_7}}}{{{T_3}}} = {\sin ^2}x \cdot {\cos ^2}x$.
Therefore, we can say that $\dfrac{{{T_3} - {T_5}}}{{{T_1}}} = \dfrac{{{T_5} - {T_7}}}{{{T_3}}}$.

Note: There are various distinct trigonometric identities. When trigonometric functions are involved in an equation then trigonometric identities are useful to solve that equation. We can use identities $\cos e{c^2}x - {\cot ^2}x = 1$ and ${\sec ^2}x - {\tan ^2}x = 1$ to solve many trigonometric problems. These identities are called Pythagorean identities.