Question

What is the relation between the sine and cosine values of angles in each quadrant? How would you use the 60° angle to find sine and cosine of 120°, 240°, and 300°? What angles could we find sine and cosine for using information for π/4 and π/6?

Answer 1

The angles, 60°, $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{\pi}{6}$ are special angles that have known trigonometric ratio values.

First part;

• The sine and cosine gives the coordinates of the tip of the radius of a unit circle as it rotates P(cos(θ), sin(θ))

Second part;

• With the knowledge of the sine and cosine of 60°, we have;
• sin(60°) = sin(120°), sin(240°) = -sin(60°), sin(300°) = -sin(60°)
• cos(120°) = -cos(60°), cos(240°) = -cos(60°), cos(300°) =  cos(60°)  

Third part;

• $\displaystyle \frac{\pi}{4}$ can be used to find the sine and cosine of $\displaystyle \frac{3 \cdot \pi}{4}$, $\displaystyle \frac{5 \cdot \pi}{4}$, and $\displaystyle \frac{7 \cdot \pi}{4}$
•  $\displaystyle \frac{\pi}{6}$, can be used to find the sine and cosine of $\displaystyle \frac{5 \cdot \pi}{6}$, $\displaystyle \frac{7 \cdot \pi}{6}$, and $\displaystyle \frac{11 \cdot \pi}{6}$

Reasons:

First Part;

Considering a unit circle with the center at the origin of the graph, we have;

The sine of the angle, θ, rotated by the radius is the vertical distance of a point P on the circle which is the location of the radius, from the horizontal axis.

The cosine of the angle, θ, is the horizontal distance of P from the vertical axis, such that we have;

The coordinates of point P = (cos(θ), sin(θ))

In the four quadrant, we have;

First Quadrant; All trigonometric ratios are positive

Second Quadrant; sine is positive

Third Quadrant; Tan is positive

Fourth Quadrant; Cosine is positive

Second part;

We have; At 120°, the point P is the same elevation from the horizontal axis, therefore;

sin(60°) = sin(120°) = 0.5·√3

However, the x-coordinate of the point P is in the negative direction, therefore, we get;

cos(120°) = -cos(60°) = -0.5

Similarly from the quadrant relationship, we have;

240° is in the third quadrant, and it is 60° below the negative horizontal line, therefore;

sin(240°) = -sin(60°) = -0.5·√3

cos(240°) = -cos(60°) = -0.5

300° is in the fourth quadrant, and it is 60° below the positive x-axis, therefore;

sin(300°) is negative and cos(300°) is positive

Which gives;

sin(300°) = -sin(60°) =  -0.5·√3

cos(300°) =  cos(60°) = 0.5

Third part;

$\displaystyle \frac{\pi}{4} =45^{\circ}$

$\displaystyle \frac{\pi}{6} =30^{\circ}$

The sine and cosine of 45° can be used to find the sine and cosine of (180° + 45°) = 225°, (360° - 45°) = 315°

Also, due to the mid location of the angle 45° on the quadrant, we have;

Another angles is the sines and cosine of (90° + 45°) = 135°

Therefore, $\displaystyle \frac{\pi}{4}$, can be used to find the sine and cosine of 135°, 225°, and 315°

$\displaystyle 135^{\circ} = \mathbf{\frac{3 \cdot \pi}{4}}$, $\displaystyle 225^{\circ} = \frac{5 \cdot \pi}{4}$, $\displaystyle 315^{\circ} = \frac{7 \cdot \pi}{4}$

Therefore,

$\displaystyle \frac{\pi}{4}$ can be used to find the sine and cosine of $\displaystyle \mathbf{\frac{3 \cdot \pi}{4}}$, $\displaystyle \mathbf{\frac{5 \cdot \pi}{4}}$, and $\displaystyle \mathbf{ \frac{7 \cdot \pi}{4}}$

Similarly, the sine and cosine of, $\displaystyle \frac{\pi}{6}$  = 30° can be used to find the sine and cosine of 150°,  210°, and 330°.

$\displaystyle 150^{\circ} = \frac{5 \cdot \pi}{6}$, $\displaystyle 210^{\circ} = \frac{7 \cdot \pi}{6}$, and $\displaystyle 330^{\circ} = \frac{11 \cdot \pi}{6}$

$\displaystyle \frac{\pi}{6}$, can be used to find the sine and cosine of $\displaystyle \mathbf{ \frac{5 \cdot \pi}{6}}$, $\displaystyle \mathbf{ \frac{7 \cdot \pi}{6}}$, and $\displaystyle \mathbf{\frac{11 \cdot \pi}{6}}$

Learn more about the sine and cosine of angles here:

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