If $A=\sin^{20}\theta +\cos^{48}\theta $ then identify the correct option.

Question

If $A=\sin^{20}\theta +\cos^{48}\theta $, then for all values $\theta$
a) $A\geq 1$
b) $ 0< A\leq 1$
c) $1<A< 3$
d) None of these

$0 \leq \sin^{20}\theta \leq 1$
$0 \leq \cos^{48}\theta \leq 1 $

So I think it is $d.)$ , but I am confused.

I look for a short and simple way.

I have studied maths upto $12$th grade.

score 7 · Answer 1 · answered Jan 24 '16 at 05:51

7

Hint

What about using $$0\leq\sin^{20}(\theta)=\sin^{18}(\theta)\sin^{2}(\theta)\leq\sin^{2}(\theta)$$ $$0\leq\cos^{48}(\theta)=\cos^{46}(\theta)\cos^{2}(\theta)\leq\cos^{2}(\theta)$$ $$0\leq\sin^{20}(\theta)+\cos^{48}(\theta)\leq\sin^{2}(\theta)+\cos^{2}(\theta)=1$$

answered Jan 24 '16 at 05:51

Claude Leibovici

260,315

1

Is option $b.)$ correct. – R K Jan 24 '16 at 05:55
It seems to be if you look at the most left and most right numbers in the last line I wrote. – Claude Leibovici Jan 24 '16 at 05:59
1

@ClaudeLeibovici, Nice method by following the options available. I was thinking of finding the actual range. – lab bhattacharjee Jan 24 '16 at 06:05
@labbhattacharjee. Thank you ! May I confess that I started thinking the same way (what is the range ?). Cheers. – Claude Leibovici Jan 24 '16 at 06:17

score 0 · Answer 2 · answered Jan 24 '16 at 05:50

0

Hint: $0 \leq \sin^2\theta \leq 1$, $0 \leq \cos^2\theta \leq 1$, and $\sin^2\theta + \cos^2\theta = 1$.

answered Jan 24 '16 at 05:50

Michael Biro

13,757

score 0 · Answer 3 · answered Jan 24 '16 at 05:52

Hint:

You should know that \begin{align} 0 \leq \sin^{20}\theta \leq \sin^2\theta \leq 1\\ 0 \leq \cos^{48}\theta \leq \cos^2\theta \leq 1 \end{align}

So

$$0\leq \sin^{20}\theta + \cos^{48}\theta \leq \sin^2\theta + \cos^2\theta $$

And the last sum is $1$.

If $A=\sin^{20}\theta +\cos^{48}\theta $ then identify the correct option.

3 Answers3