Question

the line ( y = x ) crosses the unit circle in the first and third quadrants. let ( \theta _ { 1 } ) be the angle from the positive ( x )-axis to the line in the first quadrant, and let ( \theta _ { 3 } ) be the angle from the positive ( x )-axis to the line in the third quadrant, both measured counterclockwise. the coordinates of the points of intersection of this line with the unit circle are approximately ( ( 0.707,0.707 ) ) and ( ( - 0.707,-0.707 ) ) in the first and third quadrants, respectively.
(a) what is the value of ( sin ( \theta _ { 1 } ) ) ? (round your answer to three decimal places.)
( sin ( \theta _ { 1 } ) = )
(b) what is the value of ( cos ( \theta _ { 3 } ) ) ? (round your answer to three decimal places.)
( cos ( \theta _ { 3 } ) = )
(c) suppose the line ( y = - x ) were also included on the graph. let ( \theta _ { 2 } ) be the angle from the positive ( x )-axis to the line ( y = - x ) in the second quadrant, and let ( \theta _ { 4 } ) be the angle from the positive ( x )-axis to the line ( y = - x ) in the fourth quadrant, both measured counterclockwise. use symmetry to find the following value. (round your answers to three decimal places.)
( sin ( \theta _ { 2 } ) = )
( cos ( \theta _ { 2 } ) = )
( sin ( \theta _ { 4 } ) = )
( cos ( \theta _ { 4 } ) = )

Explanation:

Step1: Find the angle in the first quadrant

The line \(y = x\) intersects the unit - circle at the point \((\cos\theta,\sin\theta)\). For the unit - circle \(x^{2}+y^{2}=1\), substituting \(y = x\) gives \(x^{2}+x^{2}=1\), i.e., \(2x^{2}=1\), \(x^{2}=\frac{1}{2}\), \(x = y=\frac{\sqrt{2}}{2}\approx0.707\). So, for \(\theta_1\) (in the first quadrant), \(\sin(\theta_1)=\frac{\sqrt{2}}{2}\approx0.707\)

Step2: Find the cosine of the angle in the third quadrant

The angle \(\theta_3\) (in the third quadrant) is related to \(\theta_1\) by \(\theta_3=\theta_1 + \pi\). Using the formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\) with \(A=\theta_1\), \(B = \pi\) (\(\cos\pi=-1\), \(\sin\pi = 0\)). \(\cos(\theta_3)=\cos(\theta_1+\pi)=\cos\theta_1\cos\pi-\sin\theta_1\sin\pi\). Since \(\cos\theta_1=\frac{\sqrt{2}}{2}\), \(\cos(\theta_3)=\frac{\sqrt{2}}{2}\times(-1)-\frac{\sqrt{2}}{2}\times0=-\frac{\sqrt{2}}{2}\approx - 0.707\)

Step3: Use symmetry for \(\theta_2\) (second - quadrant, \(y=-x\))

The angle \(\theta_2\) (in the second quadrant) is \(\theta_2=\pi-\theta_1\). Using the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\) with \(A=\pi\), \(B=\theta_1\) (\(\sin\pi = 0\), \(\cos\pi=-1\)), \(\sin(\theta_2)=\sin(\pi - \theta_1)=\sin\theta_1\cos0-\cos\theta_1\sin0=\sin\theta_1\approx0.707\), \(\cos(\theta_2)=\cos(\pi-\theta_1)=\cos\pi\cos\theta_1+\sin\pi\sin\theta_1=-\cos\theta_1\approx - 0.707\)

Step4: Use symmetry for \(\theta_4\) (fourth - quadrant, \(y = - x\))

The angle \(\theta_4\) (in the fourth quadrant) is \(\theta_4 = 2\pi-\theta_1\). Using the formula \(\cos(A - B)=\cos A\cos B+\sin A\sin B\) with \(A = 2\pi\), \(B=\theta_1\) (\(\cos(2\pi)=1\), \(\sin(2\pi)=0\)), \(\cos(\theta_4)=\cos(2\pi-\theta_1)=\cos(2\pi)\cos\theta_1+\sin(2\pi)\sin\theta_1=\cos\theta_1\approx0.707\), \(\sin(\theta_4)=\sin(2\pi-\theta_1)=\sin(2\pi)\cos\theta_1-\cos(2\pi)\sin\theta_1=-\sin\theta_1\approx - 0.707\)

Answer:

(a) \(\sin(\theta_1)=0.707\)
(b) \(\cos(\theta_3)=-0.707\)
(c) \(\sin(\theta_2)=0.707\), \(\cos(\theta_2)=-0.707\), \(\sin(\theta_4)=-0.707\), \(\cos(\theta_4)=0.707\)