Comparing with Trignometry.

Geometry Level 3

For θ ∈ ( π 4 , π 2 ) , \theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right), let A = ( cos ⁡ θ ) cos ⁡ θ , B = ( sin ⁡ θ ) cos ⁡ θ , A =( \cos \theta )^{\ \cos \theta}, B =( \sin \theta )^{\ \cos \theta}, and C = ( cos ⁡ θ ) sin ⁡ θ . C=( \cos \theta) ^{\ \sin \theta}.

Which of the following is true?

Problem Courtsey: University of South Carolina

A < B < C A<B<C A < C < B A<C<B B < A < C B<A<C B < C < A B<C<A C < A < B C<A<B

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

For θ ∈ ( π 4 , π 2 ) we have, 1 2 < cos ⁡ θ < sin ⁡ θ < 1 ( 1 ) , B A = ( sin ⁡ θ ) cos ⁡ θ ( cos ⁡ θ ) cos ⁡ θ = ( sin ⁡ θ cos ⁡ θ ) cos ⁡ θ = tan ⁡ θ cos ⁡ θ ⟹ B A > 1 Since tan ⁡ θ > 1 and cos ⁡ θ > 0 for θ ∈ ( π 4 , π 2 ) ⟹ B > A ( 2 ) C A = ( cos ⁡ θ ) sin ⁡ θ ( cos ⁡ θ ) cos ⁡ θ = cos ⁡ θ sin ⁡ θ − cos ⁡ θ ⟹ C A < 1 Since cos ⁡ θ < 1 and sin ⁡ θ − cos ⁡ θ > 0 for θ ∈ ( π 4 , π 2 ) ⟹ A > C ( 3 ) From ( 2 ) and ( 3 ) we have, C < A < B \begin{aligned}\text{For }\theta \in &\left(\frac{\pi}{4},\frac{\pi}{2}\right)\text{we have,}\\ \dfrac{1}{\sqrt{2}}&<\cos \theta<\sin \theta<1\hspace{4mm}\color{#3D99F6}(1),\\\\ \dfrac{B}{A}&=\dfrac{(\sin \theta) ^{\ \cos \theta}}{(\cos \theta) ^{\ \cos \theta}}\\ &=(\dfrac{\sin \theta}{\cos \theta})^{\ \cos \theta}\\ &=\tan \theta ^{\ \cos \theta}\\ \implies \dfrac{B}{A}&>1 \hspace{4mm}\color{#3D99F6}\small\text{Since }\tan \theta>1 \text{ and }\cos \theta>0 \text{ for }\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)\\ \implies B&>A\hspace{4mm}\color{#3D99F6}(2)\\\\ \dfrac{C}{A}&=\dfrac{(\cos \theta) ^{\ \sin \theta}}{(\cos \theta) ^{\ \cos \theta}}\\ &=\cos \theta ^{\ \sin \theta-\cos \theta}\\ \implies \dfrac{C}{A}&<1 \hspace{4mm}\color{#3D99F6}\small\text{Since }\cos \theta<1 \text{ and }\sin \theta-\cos \theta>0 \text{ for }\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)\\ \implies A&>C\hspace{4mm}\color{#3D99F6}(3)\\\\ \text{From }\color{#3D99F6}(2) &\color{#333333}\text {and }\color{#3D99F6}(3)\color{#333333}\text{ we have,}\\ &\color{#EC7300} \boxed{\color{#333333}C<A<B}\end{aligned}

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