Trigonometrical Equations 3 Question 19
20. For all $\theta$ in $[0, \pi / 2]$, show that $\cos (\sin \theta) \geq \sin (\cos \theta)$.
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Solution:
- We have, $\cos \theta+\sin \theta=\sqrt{2} \frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta$
$ =\sqrt{2} \sin \frac{\pi}{4} \cdot \cos \theta+\cos \frac{\pi}{4} \cdot \sin \theta $
$ =\sqrt{2} \sin \frac{\pi}{4}+\theta $
$\Rightarrow \cos \theta+\sin \theta \leq \sqrt{2}<\frac{\pi}{2}$
$ \begin{array}{rlrl} & \operatorname{as}, \sqrt{2} & =1.4141, \frac{\pi}{2}=1.57 \text { (approx) } \\ \Rightarrow & & \cos \theta+\sin \theta & <\frac{\pi}{2} \\ \text { Since, } & & \cos \theta & <\frac{\pi}{2}-\sin \theta \\ \Rightarrow & & \sin (\cos \theta) & <\sin \frac{\pi}{2}-\sin \theta \\ \Rightarrow & & \sin (\cos \theta) & <\cos (\sin \theta) \\ \Rightarrow & & \cos (\sin \theta) & >\sin (\cos \theta) \end{array} $
