+0  
 
+2
172
3
avatar+12 

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

DenzelMathers  Mar 28, 2017
Sort: 

3+0 Answers

 #1
avatar+76870 
+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+18609 
+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

8 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details