krit.club logo

Geometry and Trigonometry - Trigonometric Ratios and Identities

Grade 11IB

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

🔑Concepts

•

The Unit Circle: Imagine a circle with a radius of 11 unit centered at the origin (0,0)(0,0) of a Cartesian plane. For any point P(x,y)P(x, y) on this circle forming an angle θ\theta with the positive x-axis, the coordinates are defined as x=cos⁡(θ)x = \cos(\theta) and y=sin⁡(θ)y = \sin(\theta). This visual relationship demonstrates why the maximum and minimum values for sine and cosine are 11 and −1-1.

•

Radian Measure: One radian is the angle subtended at the center of a circle by an arc equal in length to the radius. Visualize an equilateral-like sector where the curved edge is the same length as the straight sides. Since a full circle is 360∘360^{\circ} or 2π2\pi radians, the conversion factor is π rad=180∘\pi \text{ rad} = 180^{\circ}.

•

The ASTC Rule (All-Sine-Tangent-Cosine): This rule helps determine the sign of trigonometric ratios across the four quadrants of a coordinate plane. In Quadrant I (top-right), all ratios are positive; in Quadrant II (top-left), only Sine is positive; in Quadrant III (bottom-left), only Tangent is positive; and in Quadrant IV (bottom-right), only Cosine is positive.

•

Exact Trigonometric Ratios: Certain angles produce clean geometric values derived from special triangles. For a 45∘−45∘−90∘45^{\circ}-45^{\circ}-90^{\circ} triangle (isosceles right triangle), the ratio is sin⁡(45∘)=12\sin(45^{\circ}) = \frac{1}{\sqrt{2}}. For a 30∘−60∘−90∘30^{\circ}-60^{\circ}-90^{\circ} triangle (half of an equilateral triangle), sin⁡(30∘)=12\sin(30^{\circ}) = \frac{1}{2} and cos⁡(30∘)=32\cos(30^{\circ}) = \frac{\sqrt{3}}{2}.

•

Pythagorean Identity: Derived from applying the Pythagorean theorem to the unit circle where x2+y2=r2x^2 + y^2 = r^2. Because x=cos⁡(θ)x = \cos(\theta), y=sin⁡(θ)y = \sin(\theta), and r=1r = 1, we get the fundamental identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1. This can be rearranged to solve for one ratio when the other is known.

•

Trigonometric Equations: Solving equations like sin⁡(θ)=k\sin(\theta) = k involves finding all possible angles within a given domain (e.g., [0,2π][0, 2\pi]). This often requires finding a principal value and then using the symmetry of the unit circle or the periodicity of the graphs to find the secondary solutions.

•

Double Angle Identities: These identities express trigonometric functions of 2θ2\theta in terms of θ\theta. They are essential for simplifying expressions and solving equations where the frequency of the wave changes. Geometrically, sin⁡(2θ)\sin(2\theta) represents the vertical coordinate when the angle is doubled on the unit circle.

📐Formulae

θ (in radians)=π180∘×θ (in degrees)\theta \text{ (in radians)} = \frac{\pi}{180^{\circ}} \times \theta \text{ (in degrees)}

l=rθl = r\theta (Arc length where θ\theta is in radians)

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

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

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta)

cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)=2cos⁡2(θ)−1=1−2sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta)

tan⁡(2θ)=2tan⁡(θ)1−tan⁡2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

💡Examples

Problem 1:

Given that cos⁡(θ)=−35\cos(\theta) = -\frac{3}{5} and π<θ<3π2\pi < \theta < \frac{3\pi}{2}, find the exact value of sin⁡(2θ)\sin(2\theta).

Solution:

  1. Identify the quadrant: Since π<θ<3π2\pi < \theta < \frac{3\pi}{2}, θ\theta is in the third quadrant (T), meaning sin⁡(θ)\sin(\theta) is negative.
  2. Use the Pythagorean identity: sin⁡2(θ)+(−35)2=1⇒sin⁡2(θ)+925=1\sin^2(\theta) + (-\frac{3}{5})^2 = 1 \Rightarrow \sin^2(\theta) + \frac{9}{25} = 1.
  3. Solve for sin⁡(θ)\sin(\theta): sin⁡2(θ)=1625⇒sin⁡(θ)=−1625=−45\sin^2(\theta) = \frac{16}{25} \Rightarrow \sin(\theta) = -\sqrt{\frac{16}{25}} = -\frac{4}{5}.
  4. Apply the double angle formula: sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)=2(−45)(−35)\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2(-\frac{4}{5})(-\frac{3}{5}).
  5. Calculate: sin⁡(2θ)=2425\sin(2\theta) = \frac{24}{25}.

Explanation:

First, we determine the missing ratio (sine) using the Pythagorean identity, ensuring we pick the negative root because the angle is in the third quadrant. Finally, we substitute both sine and cosine into the double angle formula.

Problem 2:

Solve the equation 2cos⁡2(x)+sin⁡(x)−1=02\cos^2(x) + \sin(x) - 1 = 0 for 0≤x≤2π0 \le x \le 2\pi.

Solution:

  1. Use identity to create a single trig function: 2(1−sin⁡2(x))+sin⁡(x)−1=02(1 - \sin^2(x)) + \sin(x) - 1 = 0.
  2. Expand and simplify: 2−2sin⁡2(x)+sin⁡(x)−1=0⇒−2sin⁡2(x)+sin⁡(x)+1=02 - 2\sin^2(x) + \sin(x) - 1 = 0 \Rightarrow -2\sin^2(x) + \sin(x) + 1 = 0.
  3. Multiply by −1-1 to rearrange into quadratic form: 2sin⁡2(x)−sin⁡(x)−1=02\sin^2(x) - \sin(x) - 1 = 0.
  4. Factor the quadratic: (2sin⁡(x)+1)(sin⁡(x)−1)=0(2\sin(x) + 1)(\sin(x) - 1) = 0.
  5. Solve for sin⁡(x)\sin(x): sin⁡(x)=−12\sin(x) = -\frac{1}{2} or sin⁡(x)=1\sin(x) = 1.
  6. Find values of xx in the domain: For sin⁡(x)=1\sin(x) = 1, x=π2x = \frac{\pi}{2}. For sin⁡(x)=−12\sin(x) = -\frac{1}{2}, x=7π6x = \frac{7\pi}{6} and x=11π6x = \frac{11\pi}{6}.

Explanation:

To solve an equation with both sine and cosine squared, use the Pythagorean identity to convert everything to one function (sine in this case). This results in a quadratic equation which can be factored to find the possible values for xx.