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# Geometry

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Find the value of x that makes each statement true

sin (x/3 + 10) = cos x

murod7  Mar 20, 2017
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#1
+79894
+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°

CPhill  Mar 20, 2017
#2
+7165
0

Find the value of x that makes each statement true

sin (x/3 + 10) = cos x

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

$$cosx=\sqrt{1-sin^2x}$$

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

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

$$x_1 = 12,52765316663493rad$$

$$x_2 = -6,205750411731104rad$$

!

asinus  Mar 20, 2017
#3
+18777
+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 = ?

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

heureka  Mar 21, 2017

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