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knowing that Tan(45)=1, find Sin(45) and Cos(45) with the fonamental relations

DenzelMathers  Mar 28, 2017
 #1
avatar+87293 
+1

Tan (45)  =  y / x  =     1 / 1

 

And 

 

Sin (45)   =  y / r     and   Cos (45) =  x / r    =   1 / r

 

And  r  = sqrt (1 + 1)    =  sqrt (2)

 

So

 

Sin (45)   = Cos (45)    =   1 / r =  1 / sqrt (2)   =   sqrt (2)  / 2

 

 

 

cool cool cool

CPhill  Mar 28, 2017
 #2
avatar+12 
-1

what is ''r''?

DenzelMathers  Mar 28, 2017
edited by DenzelMathers  Mar 28, 2017
 #3
avatar+19599 
+2

knowing that Tan(45)=1, find Sin(45) and Cos(45) with the fonamental relations

 

\(\tan(45^{\circ}) = 1 \\ \cot(45^{\circ}) = \frac{1}{\tan(45^{\circ})} = \frac{1}{1}=1 \)

 

Formula:

\(\begin{array}{|rcll|} \hline \sin^2(x)+\cos^2(x) &=& 1 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \sin^2(45^{\circ})+\cos^2(45^{\circ}) &=& 1 \quad & | \quad : \cos^2(45^{\circ}) \\ \frac{ \sin^2(45^{\circ})+\cos^2(45^{\circ}) } {\cos^2(45^{\circ})} &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ \frac{ \sin^2(45^{\circ}) }{\cos^2(45^{\circ})} + \frac{ \cos^2(45^{\circ}) } {\cos^2(45^{\circ})} &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ \tan^2(45^{\circ}) + 1 &=& \frac{ 1 } {\cos^2(45^{\circ})} \quad & | \quad \tan(45^{\circ}) = 1 \\ 1 + 1 &=& \frac{ 1 } {\cos^2(45^{\circ})} \\ 2 &=& \frac{ 1 } {\cos^2(45^{\circ})} \quad & | \quad \text{square root both sides} \\ \sqrt{2} &=& \frac{ 1 } {\cos(45^{\circ})} \\ \mathbf{ \frac{1}{\sqrt{2}} } & \mathbf{=} & \mathbf{ \cos(45^{\circ}) } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \sin^2(45^{\circ})+\cos^2(45^{\circ}) &=& 1 \quad & | \quad : \sin^2(45^{\circ}) \\ \frac{ \sin^2(45^{\circ})+\cos^2(45^{\circ}) } {\sin^2(x)} &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ \frac{ \sin^2(45^{\circ}) }{\sin^2(45^{\circ})} + \frac{ \cos^2(45^{\circ}) } {\sin^2(45^{\circ})} &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ 1+ \cot^2(45^{\circ}) &=& \frac{ 1 } {\sin^2(45^{\circ})} \quad & | \quad \cot(45^{\circ}) = 1 \\ 1+ 1 &=& \frac{ 1 } {\sin^2(45^{\circ})} \\ 2 &=& \frac{ 1 } {\sin^2(45^{\circ})} \quad & | \quad \text{square root both sides} \\ \sqrt{2} &=& \frac{ 1 } {\sin(45^{\circ})} \\ \mathbf{ \frac{1}{\sqrt{2}} } & \mathbf{=} & \mathbf{ \sin(45^{\circ}) } \\ \hline \end{array}\)

 

laugh

heureka  Mar 29, 2017

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