+0  
 
+1
318
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Find the value of x that makes each statement true

 

 

 

sin (x/3 + 10) = cos x

murod7  Mar 20, 2017
 #1
avatar+87334 
+3

Since sine and cosine are co-functions, we have....

 

x/3 + 10 + x  =   90      subtract 10 from both sides and simplify

 

(4/3)x = 80       multiply  both sides by 3/4

 

x = (3/4)* 80    =   60°

 

 

cool cool cool

CPhill  Mar 20, 2017
 #2
avatar+7348 
0

Find the value of x that makes each statement true

 

sin (x/3 + 10) = cos x

 

\(sin(\frac{x}{3}+10)=cosx\)

 

\(cosx=\sqrt{1-sin^2x}\) 

 

\(sin(\frac{x}{3}+10)=\sqrt{1-sin^2x}\)

 

\(sin^2(\frac{x}{3}+10)=1-sin^2x\)

 

\(x_1 = 12,52765316663493rad\)

 

\(x_2 = -6,205750411731104rad\)

 

laugh  !

asinus  Mar 20, 2017
 #3
avatar+19655 
+1

Find the value of x that makes each statement true

sin (x/3 + 10) = cos x


1. x1 = ?

\(\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \sin(90^{\circ}-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}-x \\ x+\frac{x}{3}+10^{\circ} &=& 90^{\circ} \\ x+\frac{x}{3} &=& 80^{\circ} \\ x\cdot (1+\frac{1}{3}) &=& 80^{\circ} \\ x\cdot (\frac{4}{3}) &=& 80^{\circ} \\ x &=& 80^{\circ}\cdot \frac{3}{4} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}\)

 

2. x2 = ?

\(\begin{array}{|rcll|} \hline \sin(\frac{x}{3}+10^{\circ}) &=& \cos(x) \quad & | \quad \cos(x) = \cos(-x) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \cos(-x) \quad & | \quad \cos(-x) = \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}-(-x)) \\ \sin(\frac{x}{3}+10^{\circ}) &=& \sin(90^{\circ}+x) \\ \frac{x}{3}+10^{\circ} &=& 90^{\circ}+x \\ x-\frac{x}{3} &=& -80^{\circ} \\ x\cdot (1-\frac{1}{3}) &=& -80^{\circ} \\ x\cdot (\frac{2}{3}) &=& -80^{\circ} \\ x &=& -80^{\circ}\cdot \frac{3}{2} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-120^{\circ} } \\ \hline \end{array}\)

 

laugh

heureka  Mar 21, 2017

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