krit.club logo

Geometry and Trigonometry - The unit circle and radian measure

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Unit Circle Definition: The unit circle is a circle with a radius of 11 unit centered at the origin (0,0)(0, 0) on a Cartesian coordinate system. It serves as the foundation for defining trigonometric functions for any angle. Visually, any point P(x,y)P(x, y) on the circle's boundary satisfies the equation x2+y2=1x^2 + y^2 = 1.

•

Radian Measure: A radian is an alternative unit for measuring angles, defined as the angle subtended at the center of a circle by an arc whose length is equal to the radius. Visually, if you take the radius of a circle and wrap it along the circumference, the angle created is 11 radian. A full rotation of 360∘360^\circ is equivalent to 2π2\pi radians.

•

Coordinates as Trigonometric Ratios: For any angle θ\theta in standard position (starting from the positive x-axis and rotating counter-clockwise), the intersection point (x,y)(x, y) on the unit circle has coordinates x=cos⁡(θ)x = \cos(\theta) and y=sin⁡(θ)y = \sin(\theta). Visually, as the point PP moves around the circle, the vertical distance from the x-axis represents the sine, and the horizontal distance from the y-axis represents the cosine.

•

Tangent on the Unit Circle: The tangent of an angle θ\theta is defined as the ratio sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)}, which corresponds to yx\frac{y}{x} on the unit circle. Visually, tan⁡(θ)\tan(\theta) can be represented as the length of a segment on a line tangent to the circle at (1,0)(1, 0) that is intercepted by the terminal side of the angle.

•

The CAST Diagram and Quadrants: The signs of trigonometric ratios depend on the quadrant in which the angle's terminal side lies. In Quadrant I (0<θ<π20 < \theta < \frac{\pi}{2}), All are positive; in Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), only Sine is positive; in Quadrant III (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), only Tangent is positive; in Quadrant IV (3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi), only Cosine is positive.

•

Special Angles in Radians: Common angles from geometry are frequently represented in radians: 30∘=π630^\circ = \frac{\pi}{6}, 45∘=π445^\circ = \frac{\pi}{4}, 60∘=π360^\circ = \frac{\pi}{3}, and 90∘=π290^\circ = \frac{\pi}{2}. Visually, these correspond to specific symmetric points on the unit circle, such as (32,12)(\frac{\sqrt{3}}{2}, \frac{1}{2}) for π6\frac{\pi}{6}.

•

Arc Length and Sector Area: When using radians, the formulas for the geometry of a circle are simplified. The arc length ss is the portion of the circumference, and the sector area AA is the 'slice' of the circle. Visually, as the angle θ\theta increases, both the length of the 'crust' (arc) and the 'area of the slice' (sector) increase proportionally to the radius.

📐Formulae

Degrees to Radians: θrad=θdeg×π180\theta_{rad} = \theta_{deg} \times \frac{\pi}{180}

Radians to Degrees: θdeg=θrad×180π\theta_{deg} = \theta_{rad} \times \frac{180}{\pi}

Unit Circle Identity: cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1

Tangent Ratio: tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

Arc Length: s=rθs = r\theta (where θ\theta is in radians)

Area of a Sector: A=12r2θA = \frac{1}{2}r^2\theta (where θ\theta is in radians)

💡Examples

Problem 1:

Convert an angle of 150∘150^\circ to radians and calculate the length of the arc it subtends in a circle with a radius of 66 cm.

Solution:

Step 1: Convert degrees to radians using the formula θ×π180\theta \times \frac{\pi}{180}. θ=150×π180=15π18=5π6 radians\theta = 150 \times \frac{\pi}{180} = \frac{15\pi}{18} = \frac{5\pi}{6} \text{ radians} Step 2: Use the arc length formula s=rθs = r\theta. s=6×5π6s = 6 \times \frac{5\pi}{6} s=5π≈15.71 cms = 5\pi \approx 15.71 \text{ cm}

Explanation:

To find arc length, the angle must first be converted from degrees to radians. Multiplying the resulting radian measure by the radius gives the physical length of the curve along the circle's edge.

Problem 2:

Given a point PP on the unit circle in Quadrant II where sin⁡(θ)=45\sin(\theta) = \frac{4}{5}, find the value of cos⁡(θ)\cos(\theta) and tan⁡(θ)\tan(\theta).

Solution:

Step 1: Use the identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1. cos⁡2(θ)+(45)2=1\cos^2(\theta) + (\frac{4}{5})^2 = 1 cos⁡2(θ)+1625=1\cos^2(\theta) + \frac{16}{25} = 1 cos⁡2(θ)=1−1625=925\cos^2(\theta) = 1 - \frac{16}{25} = \frac{9}{25} Step 2: Determine the sign of cos⁡(θ)\cos(\theta). In Quadrant II, the x-coordinate (cosine) is negative. cos⁡(θ)=−925=−35\cos(\theta) = -\sqrt{\frac{9}{25}} = -\frac{3}{5} Step 3: Calculate tan⁡(θ)\tan(\theta) using the ratio sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)}. tan⁡(θ)=4/5−3/5=−43\tan(\theta) = \frac{4/5}{-3/5} = -\frac{4}{3}

Explanation:

We use the Pythagorean identity inherent in the unit circle to find the magnitude of the cosine value. The quadrant information is then critical to choosing the correct negative sign for the cosine and tangent values.