Trigonometric Functions — MATHS STD 11 Science — Question

1. $\sin^6\text{x}+\cos^6\text{x}=$ $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$
2. $|\sin\text{x}-\cos\text{x}|=$ $\sqrt{2-\text{a}^2}.$

1. $\sin^6\text{x}+\cos^6\text{x}=(\sin^2\text{x})^3+(\cos^2\text{x})3$

$=(\sin^2\text{x}+\cos^2\text{x})^3-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})$

$=(1)^3-3\Big(\frac{\text{a}^2-1}{2}\Big)^2\cdot1=1-\frac{3(\text{a}^2-1)^2}{4}$

$=\frac{1}{4}[4-3(\text{a}^2-1)^2]$

Hence ,the value of the filler is $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$

2. $|\sin\text{x}-\cos\text{x}|^2=\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}$

$=1-2\Big(\frac{\text{a}^2-1}{2}\Big)=1-(\text{a}^2-1)=1-\text{a}^2+1$

$=2-\text{a}^2$

$\therefore|\sin\text{x}-\cos\text{x}|=\sqrt{2-\text{a}^2}$ $[\therefore|\sin\text{x}-\cos\text{x}|>0]$

Hence , the value of the filler is $\sqrt{2-\text{a}^2}\cdot$