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1) \(csc^3(w)=\frac{cot^3(w)+cot(w)}{cos(w)}\)

AdamTaurus  Mar 28, 2018
edited by AdamTaurus  Mar 28, 2018
edited by AdamTaurus  Mar 28, 2018
edited by AdamTaurus  Mar 29, 2018
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2+0 Answers

 #1
avatar+6943 
+2

Note that...

 

\(\sin^2(w)+\cos^2(w)=1\quad\rightarrow\quad \frac{\sin^2(w)}{\sin^2(w)}+\frac{\cos^2(w)}{\sin^2(w)}=\frac{1}{\sin^2(w)}\quad\rightarrow\quad 1+\cot^2(w)=\csc^2(w)\)

 

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\(\begin{array}\ \csc^3(w)&=&\frac{\cot^3(w)+\cot(w)}{\cos(w)}\\~\\ &=&\frac{\cot(w)(\cot^2(w)+1)}{\cos(w)}\\~\\ &=&\frac{\cot(w)(1+\cot^2(w))}{\cos(w)}\\~\\ &=&\frac{\cot(w)\csc^2(w)}{\cos(w)}\\~\\ &=&\frac{1}{\cos(w)}\cdot\cot(w)\cdot\csc^2(w)\\~\\ &=&\frac{1}{\cos(w)}\cdot\frac{\cos(w)}{\sin(w)}\cdot\csc^2(w)\\~\\ &=&\frac{1}{\sin(w)}\cdot\csc^2(w)\\~\\ &=&\csc(w)\cdot\csc^2(w)\\~\\ &=&\csc^3(w) \end{array} \)

hectictar  Mar 29, 2018
 #2
avatar+19207 
+2

Verify (Prove) Each Identity

csc^3(w)=\frac{cot^3(w)+cot(w)}{cos(w)}

\(\displaystyle csc^3(w)=\frac{cot^3(w)+cot(w)}{cos(w)} \)

 

\(\begin{array}{|rcll|} \hline \csc^3(w) &=& \dfrac{\cot^3(w)+\cot(w)} {\cos(w)} \\\\ &=&\dfrac{\cot(w)\cdot \left( \cot^2(w)+1 \right)}{\cos(w)} \quad & | \quad \cot(w) = \dfrac{\cos(w)}{\sin(w)} \\\\ &=&\dfrac{\dfrac{\cos(w)}{\sin(w)}\cdot \left( \dfrac{\cos^2(w)}{\sin^2(w)}+1 \right)}{\cos(w)} \\\\ &=&\dfrac{\dfrac{\cos(w)}{\sin(w)}\cdot \left( \dfrac{\cos^2(w)+\sin^2(w)}{\sin^2(w)} \right)}{\cos(w)} \quad & | \quad \cot^2(w) + \sin^2(w) =1 \\\\ &=&\dfrac{\dfrac{\cos(w)}{\sin(w)}\cdot \left( \dfrac{1}{\sin^2(w)} \right)}{\cos(w)} \\\\ &=&\dfrac{\dfrac{\cos(w)}{\sin^3(w)}}{\cos(w)} \\\\ &=& \dfrac{\cos(w)} {\sin^3(w)\cos(w)} \\\\ &=& \dfrac{1} {\sin^3(w)} \quad & | \quad \dfrac{1}{\sin(w)} = \csc(w) \\\\ &=& \csc^3(w) \\ \hline \end{array}\)

 

 

laugh

heureka  Mar 29, 2018

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