krit.club logo

Geometry and Trigonometry - Trigonometric identities and equations

Grade 10IB

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

🔑Concepts

•

The Unit Circle: A circle centered at the origin (0,0)(0,0) with a radius of 11 unit. For any point (x,y)(x, y) on the circle making an angle θ\theta with the positive x-axis, the coordinates are defined as x=cos⁡θx = \cos\theta and y=sin⁡θy = \sin\theta. Visually, this creates a right-angled triangle within the circle where the hypotenuse is 11, the opposite side is sin⁡θ\sin\theta, and the adjacent side is cos⁡θ\cos\theta.

•

The CAST Rule (ASTC): This quadrant-based rule determines the sign of trigonometric ratios. In the first quadrant (All), sin⁡\sin, cos⁡\cos, and tan⁡\tan are positive. In the second quadrant (Sine), only sin⁡\sin is positive. In the third quadrant (Tangent), only tan⁡\tan is positive. In the fourth quadrant (Cosine), only cos⁡\cos is positive. Visually, this follows a counter-clockwise path starting from the top-right quadrant.

•

Fundamental Trigonometric Identities: These are equations that are true for all values of θ\theta. The quotient identity states tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is derived from the Pythagorean theorem x2+y2=r2x^2 + y^2 = r^2 applied to the unit circle.

•

Reference Angles and Symmetry: A reference angle θref\theta_{ref} is the acute angle between the terminal side of an angle and the x-axis. Due to the symmetry of the unit circle, sin⁡(180∘−θ)=sin⁡θ\sin(180^\circ - \theta) = \sin\theta (horizontal reflection) and cos⁡(360∘−θ)=cos⁡θ\cos(360^\circ - \theta) = \cos\theta (vertical reflection). These properties allow us to find multiple solutions for trigonometric equations within a given range.

•

Solving Trigonometric Equations: To solve equations like sin⁡θ=k\sin\theta = k, first find the primary value (inverse sine). Then, use the unit circle or the CAST rule to find the second solution within the domain (usually 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ). Visually, a horizontal line y=ky=k intersecting the unit circle or the sine wave graph shows why there are typically two solutions per cycle.

•

Trigonometric Graphs: The sine and cosine functions produce periodic waves. The sine graph y=sin⁡θy = \sin\theta starts at (0,0)(0,0) and reaches a peak at 90∘90^\circ, while the cosine graph y=cos⁡θy = \cos\theta starts at its maximum (0,1)(0,1). Both have a period of 360∘360^\circ (or 2π2\pi radians) and an amplitude of 11, oscillating between the horizontal lines y=1y=1 and y=−1y=-1.

📐Formulae

tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}

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

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

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

sin⁡(180∘−θ)=sin⁡θ\sin(180^\circ - \theta) = \sin\theta

cos⁡(360∘−θ)=cos⁡θ\cos(360^\circ - \theta) = \cos\theta

tan⁡(180∘+θ)=tan⁡θ\tan(180^\circ + \theta) = \tan\theta

💡Examples

Problem 1:

Solve the equation 2cos⁡θ+1=02\cos\theta + 1 = 0 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.

Solution:

  1. Isolate cos⁡θ\cos\theta: 2cos⁡θ=−12\cos\theta = -1 cos⁡θ=−12\cos\theta = -\frac{1}{2}

  2. Find the reference angle α\alpha by taking the inverse of the positive value: α=cos⁡−1(12)=60∘\alpha = \cos^{-1}(\frac{1}{2}) = 60^\circ

  3. Determine the quadrants where cos⁡\cos is negative: Quadrant II and Quadrant III.

  4. Calculate the solutions in those quadrants: Quadrant II: θ=180∘−60∘=120∘\theta = 180^\circ - 60^\circ = 120^\circ Quadrant III: θ=180∘+60∘=240∘\theta = 180^\circ + 60^\circ = 240^\circ

Final Answer: θ=120∘,240∘\theta = 120^\circ, 240^\circ

Explanation:

This problem involves isolating the trigonometric ratio and then using the reference angle and the CAST rule to find all possible values of θ\theta within the specified domain where the cosine value is negative.

Problem 2:

Given that sin⁡θ=45\sin\theta = \frac{4}{5} and 90∘<θ<180∘90^\circ < \theta < 180^\circ, find the exact value of cos⁡θ\cos\theta and tan⁡θ\tan\theta.

Solution:

  1. Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: (45)2+cos⁡2θ=1(\frac{4}{5})^2 + \cos^2\theta = 1 1625+cos⁡2θ=1\frac{16}{25} + \cos^2\theta = 1

  2. Solve for cos⁡2θ\cos^2\theta: cos⁡2θ=1−1625=925\cos^2\theta = 1 - \frac{16}{25} = \frac{9}{25} cos⁡θ=±925=±35\cos\theta = \pm\sqrt{\frac{9}{25}} = \pm\frac{3}{5}

  3. Determine the sign based on the quadrant: Since 90∘<θ<180∘90^\circ < \theta < 180^\circ (Quadrant II), cos⁡\cos must be negative. cos⁡θ=−35\cos\theta = -\frac{3}{5}

  4. Calculate tan⁡θ\tan\theta: tan⁡θ=sin⁡θcos⁡θ=4/5−3/5=−43\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{4/5}{-3/5} = -\frac{4}{3}

Final Answer: cos⁡θ=−35,tan⁡θ=−43\cos\theta = -\frac{3}{5}, \tan\theta = -\frac{4}{3}

Explanation:

This example uses the Pythagorean identity to find a missing ratio and then applies knowledge of quadrants to select the correct sign for the results.