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[Solved] Trigonometric functions

 
Trigonometry 1 & 2
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June 15, 2021 2:16 pm
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(@armor)
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June 15, 2021 2:16 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Let   \small sinx+cosx=u,   then    \small sinxcosx=\frac{u^2-1}{2},   and    \small u=\sqrt{2}sin(x+\frac{\Pi }{4}).

When   \small x=\frac{\Pi }{4}, u_{max}=\sqrt{2}.    

When \small x=-\frac{\Pi }{6},   \small u  reaches the minimum value. 

Since \small u> 0,  then  \small u_{min}=\sqrt{\frac{2-\sqrt{3}}{2}}=\frac{\sqrt{3}-1}{2},   then  \small u\epsilon [\frac{\sqrt{3}-1}{2},\sqrt{2}].

Since \small y=(sinx+1)(cosx+1)=sinx+cosx+sinxcosx+1=\frac{1}{2}(u+1)^2,  and  \small y isincreasing on the interval 

\small [\frac{\sqrt{3}-1}{2},\sqrt{2}],   therefore,  \small y_{min}=\frac{2+\sqrt{3}}{4}, y_{max}=\frac{3+2\sqrt{2}}{2}.

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