Geometry

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°

Find the value of x that makes each statement true

sin (x/3 + 10) = cos x

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

$cosx=\sqrt{1-sin^2x}$ 

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

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

$x_1 = 12,52765316663493rad$

$x_2 = -6,205750411731104rad$

  !

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 = ?

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