Solve the Following Question.(5 Marks)

Question 15 Marks

If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$

Answer

We Have $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x}\cdots(\text{i})$
To show: $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
$\text{L.H.S}=\frac{\text{T}_3-\text{T}_5}{\text{T}_1}$
$=\frac{(\sin^3\text{x}+\cos^3\text{x})-(\sin^5\text{x}+\cos^5\text{x})}{\sin\text{x}+\cos\text{x}} [$Substituting the value of $T_3, T_5$ and $T_1$ From $(i)]$
$=\frac{\sin^3\text{x}-\sin^5\text{x}+\cos^3\text{x}-\cos^5\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}-(1-\sin^2\text{x})+\cos^3\text{x}(1-\cos^2\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}\cos^2\text{x}+\cos^3\text{x}\sin^2\text{x}}{\sin\text{x}+\cos\text{x}}$ $\Big[\because1-\sin^2\text{x}=\cos^2\text{x}\text{ and }1-\cos^2\text{x}=\sin^2\text{x}\Big]$
$=\frac{\sin^2\text{x}\cos^2\text{x}+(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{R.H.S}=\frac{\sin^5\text{x}+\cos^5\text{x}-(\sin^7\text{x}+\cos^7\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}-\sin^7\text{x}+\cos^5\text{x}-\cos^7\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}(1-\sin^2\text{x})+\cos^5\text{x}(1-\cos^2\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}+\cos^5\text{x}\sin^2\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^3\text{x})}{\sin^2\text{x}+\cos^2\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{L.H.S=R.H.S }$
$\text{Proved}$

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Question 25 Marks

If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$6\text{T}_{10}-15\text{T}_8+10\text{T}_6-1=0$

Answer

$=-6\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})+6\sin^4\cos^4\text{x}\\\ \ \ +9\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})=6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ +2\sin^2\text{x}\cos^2\text{x }2\sin^2\text{x}\cos^2)\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $($Adding and subtracting $2\sin^2\text{x}\cos^2\text{x})$$$
$=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x})\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}(1-2\sin^2\text{x}\cos^2\text{x})+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=3\sin^2\text{x}\cos^2\text{x}-6\sin^4\text{x}\cos^4+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$
$=0$
$=\text{R.H.S}$
$\text{Proved}.$

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Question 35 Marks

If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$2 \text{T}_6 - 3\text{ T}_4 + 1 = 0$

Answer

$\text{L.H.S}=2\text{T}_6-3\text{T}_4+1$
$=2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1$
$=2((\sin^2\text{x})^3+(\cos^2\text{x})^3-2(\sin^2\text{x})^2+(\cos^2\text{x})^2)+1$
$=2((\sin^2\text{x}+\cos^2\text{x})(\sin^2\text{x})^2+(\cos^2\text{x})^2-(\sin^2\text{x}\cos^2\text{x}))\\\ \ \ -3((\sin^2\text{x})^2+(\cos^2\text{x})^2+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})=1 [$ Using $a^3 + b^3 =(a + b)(a^2 + b^2 - ab)$ and adding and subtracting $2\sin^2\text{x}\cos^2\text{x}\big]$
$=2((\sin^2\text{x}+\cos^2\text{x})^2-3\sin^2\text{x}\cos^2\text{x})-3(1-2\sin^2\text{x}\cos^2\text{x})+1$
$=2(1-3\sin^2\text{x}\cos^2\text{x})-3\sin^2\text{x}\cos^2\text{x}+1$
$=2-6\sin^2\text{x}\cos^2\text{x}-2+6\sin^2\text{x}\cos^2\text{x}$
$=0$
$=\text{R.H.S}$
$\text{Proved}.$

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