Prove the following trigonometric identities. If T_n= ^n + ^n , porve that T_3-T_5 T_1 = T_5-T_7 T_3 .

In the given question, we are given T_n= ^n + ^n We need to prove T_3-T_5 T_1 = T_5-T_7 T_3 Here L.H.S is L.H.S= T_3-T_5 T_1 =( ^3 + ^3 )-( ^5 + ^5 ) ( + ) Now, solving the L.H.S, we get ( ^3 + ^3 )-( ^5 + ^5 ) ( + ) = ^3 - ^3 + ^3 - ^3 + = ^3 (1- ^2 )+ ^3 (1- ^2 ) + Further using the property ^2 + ^2 =1, we get ^2 =1- ^2 ^2 =1- ^2 = ^3 (1- ^2 )+ ^3 (1- ^2 ) + = ^3 ^2 + ^3 ^2 + = ^2 ^2 ( + ) + = ^2 ^2 Now, solving the R.H.S, we get R.H.S= T_5-T_7 T_3 =( ^5 + ^5 )-( ^7 + ^7 ) ( ^3 + ^3 ) So, ( ^5 + ^5 )-( ^7 + ^7 ) ( ^3 + ^3 ) = ^5 - ^7 + ^5 - ^7 ^3 + ^3 = ^5 (1- ^2 )+ ^5 (1- ^2 ) ^3 + ^3 Furhter using the property ^2 + ^2 =1, we get, cos^2 =1- ^2 ^2 =1- ^2 So, = ^5 (1- ^2 )+ ^5 (1- ^2 ) ^3 + ^3 = ^5 ^2 + ^5 ^2 ^3 + ^3 = ^2 ^2 ( ^3 + ^3 ) ^3 + ^3 = ^2 ^2 L.H.S=R.H.S

Prove the following trigonometric identities. If T_n= ^n + ^n , p…

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Question

Prove the following trigonometric identities.
If $\text{T}_\text{n}=\sin^\text{n}\theta+\cos^\text{n}\theta,$ porve that $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}.$

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Answer

In the given question, we are given $\text{T}_\text{n}=\sin^\text{n}\theta+\cos^\text{n}\theta$
We need to prove $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
Here L.H.S is
$\text{L.H.S}=\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{(\sin^3\theta+\cos^3\theta)-(\sin^5\theta+\cos^5\theta)}{(\sin\theta+\cos\theta)}$
Now, solving the L.H.S, we get
$\frac{(\sin^3\theta+\cos^3\theta)-(\sin^5\theta+\cos^5\theta)}{(\sin\theta+\cos\theta)}=\frac{\sin^3\theta-\sin^3\theta+\cos^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}$
$=\frac{\sin^3\theta(1-\sin^2\theta)+\cos^3\theta(1-\cos^2\theta)}{\sin\theta+\cos\theta}$
Further using the property $\sin^2\theta+\cos^2\theta=1,$ we get
$\cos^2\theta=1-\sin^2\theta$
$\sin^2\theta=1-\cos^2\theta$
$=\frac{\sin^3\theta(1-\sin^2\theta)+\cos^3\theta(1-\cos^2\theta)}{\sin\theta+\cos\theta}=\frac{\sin^3\theta\cos^2\theta+\cos^3\theta\sin^2\theta}{\sin\theta+\cos\theta}$
$=\frac{\sin^2\theta\cos^2\theta(\sin\theta+\cos\theta)}{\sin\theta+\cos\theta}$
$=\sin^2\theta\cos^2\theta$
Now, solving the R.H.S, we get
$\text{R.H.S}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}=\frac{(\sin^5\theta+\cos^5\theta)-(\sin^7\theta+\cos^7\theta)}{(\sin^3\theta+\cos^3\theta)}$
So,
$\frac{(\sin^5\theta+\cos^5\theta)-(\sin^7\theta+\cos^7\theta)}{(\sin^3\theta+\cos^3\theta)}=\frac{\sin^5\theta-\sin^7\theta+\cos^5\theta-\cos^7\theta}{\sin^3\theta+\cos^3\theta}$
$=\frac{\sin^5\theta(1-\sin^2\theta)+\cos^5\theta(1-\cos^2\theta)}{\sin^3\theta+\cos^3\theta}$
Furhter using the property $\sin^2\theta+\cos^2\theta=1,$ we get,
$\text{cos}^2\theta=1-\sin^2\theta$
$\sin^2\theta=1-\cos^2\theta$
So,
$=\frac{\sin^5\theta(1-\sin^2\theta)+\cos^5\theta(1-\cos^2\theta)}{\sin^3\theta+\cos^3\theta}=\frac{\sin^5\theta\cos^2\theta+\cos^5\theta\sin^2\theta}{\sin^3\theta+\cos^3\theta}$
$=\frac{\sin^2\theta\cos^2\theta(\sin^3\theta+\cos^3\theta)}{\sin^3\theta+\cos^3\theta}$
$=\sin^2\theta\cos^2\theta$
$\therefore\ \text{L.H.S}=\text{R.H.S}$