Answer
Verified463.8k+ views
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.
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.
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE