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[Solved] Comparison of trigonometric functions

 
Trigonometry 1 & 2
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June 15, 2021 2:14 pm
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(@armor)
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June 15, 2021 2:15 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
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Let  \small arcsin\frac{1}{3}=\alpha ,   then     \small sin\alpha =\frac{1}{3}< \frac{1}{2},  thus    \small 0< \alpha < \frac{\Pi }{6}.

Let \small arctan\sqrt{2}=\beta ,  then    \small tan\beta =\sqrt{2}.    Since    \small 1< \sqrt{2}< \sqrt{3},     then       \small \frac{\Pi }{4}< \beta < \frac{\Pi }{3}.

Let \small arccos\frac{3}{4}=\gamma ,   then   \small cos\gamma =\frac{3}{4}.   Since     \small \frac{\sqrt{2}}{2}< \frac{3}{4}< \frac{\sqrt{3}}{2},   then \small \frac{\Pi }{6}< \gamma < \frac{\Pi }{4}.

As a conclusion,  \small \alpha < \gamma < \beta ,  therefore,  \small arccsin\frac{1}{3}< arccos\frac{3}{4}< arctan\sqrt{2}.

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