Question

If A=cos²B+sin⁴B
Then p.v that for all values of B,3/4<=A<=1.

Answer 1

Given,

$\longrightarrow A=\cos^2B+\sin^4B$

Since $\sin^2B+\cos^2B=1,$

$\longrightarrow A=\cos^2B+(1-\cos^2B)^2$

$\longrightarrow A=\cos^2B+1-2\cos^2B+\cos^4B$

$\longrightarrow A=\cos^4B-\cos^2B+1$

$\longrightarrow A=\cos^2B(\cos^2B-1)+1$

$\longrightarrow A=1-\sin^2B\cos^2B$

$\longrightarrow A=1-\dfrac{1}{4}(2\sin B\cos B)^2$

Since $2\sin B\cos B=\sin(2B),$

$\longrightarrow A=1-\dfrac{1}{4}\sin^2(2B)$

We know that $\sin x\in[-1,\ 1].$ So,

$\longrightarrow\sin(2B)\in[-1,\ 1]$

Squaring, we get,

$\longrightarrow\sin^2(2B)\in[0,\ 1]$

Dividing by 4,

$\longrightarrow\dfrac{1}{4}\sin^2(2B)\in\left[0,\ \dfrac{1}{4}\right]$

Multiplying by -1, (note the interval limit change)

$\longrightarrow-\dfrac{1}{4}\sin^2(2B)\in\left[-\dfrac{1}{4},\ 0\right]$

Adding 1,

$\longrightarrow1-\dfrac{1}{4}\sin^2(2B)\in\left[\dfrac{3}{4},\ 1\right]$

That is,

$\longrightarrow\underline{\underline{A\in\left[\dfrac{3}{4},\ 1\right]}}\quad\quad\forall\,B\in\mathbb{R}$

Hence Proved!