Question

Let $\theta \in [0,\frac{\pi }{2}]$. Which one of the following is true ?

• A. ${\sin }^2\theta >{\cos }^2\theta$
• B. ${\sin }^2\theta <{\cos }^2\theta$
• C. $\sin \theta >\cos \theta$
• D. $\cos \theta >\sin \theta$
• E. $\sin \theta +\cos \theta \le \sqrt{2}$

Answer 1

The correct option is E $\sin \theta +\cos \theta \le \sqrt{2}$

$\cos \theta >\sin \theta ⋎\theta \in (0,\frac{\pi }{4})$ and

$\sin \theta <\cos \theta ⋎\theta \in (\frac{\pi }{4},\frac{\pi }{2})$
So options $A,B,C$ and $D$ are all incorrect

$-\sqrt{a^2+b^2}\le a\sin \theta +b\cos \theta \le \sqrt{a^2+b^2}\Rightarrow -\sqrt{1^2+1^2}\le \sin \theta +\cos \theta \le \sqrt{1^2+1^2}\Rightarrow \sin \theta +\cos \theta \le \sqrt{2}$

So option $E$ is correct.