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[Solved] Trigonometry question

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 2:59 pm
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(@armor)
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June 15, 2021 3:00 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

(1)  

Proof: Since   \small 2f(-sinx)+3f(sinx)=4sinxcosx     (I), 

 substitute \small -sinx   into (I)  to obtain \small 2f(sinx)+3f(-sinx)=-4sinxcosx     (II)

\small (I)+(II)\Rightarrow 5f(sinx)+5f(-sinx)=0\Rightarrow f(sinx)=-f(-sinx),   therefore  \small f(x)  is an odd function.

(2)

Solution: \small (I)-(II)\Rightarrow f(sinx)-f(-sinx)=8sinxcosx. 

Since \small f(sinx)=-f(-sinx)\Rightarrow 2f(sinx)=8sinxcosx,  then  \small f(sinx)=4sinx\sqrt{1-sin^2x}, (-1\leq x\leq 1).

Therefore, \small f(x)=4x\sqrt{1-x^2}, (-1\leq x\leq 1).

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