Question

If $A=si{n}^8\theta +co{s}^{14}\theta ,$ then for all real values of $\theta$

• A.
• B. 0 <
• C. 1 <
• D.

Answer 1

The correct option is B 0 <

Given that $A=si{n}^8\theta +co{s}^{14}\theta ,$
we know that $(si{n}^2\theta {)}^4\le si{n}^2\theta and(co{s}^2\theta {)}^7\le co{s}^2\theta$
Adding, we that $si{n}^8\theta +co{s}^{14}\theta \le 1$
or 0 < $si{n}^8\theta +co{s}^{14}\theta \le 1.$ Hence 0 < $A\le 1.$