krit.club logo

Trigonometry - Trigonometric Functions and their Graphs

Grade 11ICSE

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

🔑Concepts

•

Angle Measurement and Radians: Angles can be measured in degrees or radians. One radian is the angle subtended at the center of a circle by an arc length equal to the radius. The relationship is defined as π\pi radians =180∘= 180^{\circ}. Visually, a full rotation is 2π2\pi radians, representing the entire circumference of a unit circle.

•

Trigonometric Functions on the Unit Circle: In a unit circle with radius r=1r = 1, any point P(x,y)P(x, y) on the circumference corresponding to an angle θ\theta is represented as (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta). This defines sin⁡θ\sin \theta as the vertical coordinate (height) and cos⁡θ\cos \theta as the horizontal coordinate. The tangent is the ratio yx\frac{y}{x}, representing the slope of the terminal side.

•

The ASTC Rule (Signs of Functions): The coordinate plane is divided into four quadrants which determine the sign of trig functions. In Quadrant I (All), all functions are positive. In Quadrant II (Silver/Sin), only sin⁡\sin and csc⁡\csc are positive. In Quadrant III (Tea/Tan), only tan⁡\tan and cot⁡\cot are positive. In Quadrant IV (Cups/Cos), only cos⁡\cos and sec⁡\sec are positive.

•

Domain and Range: For f(x)=sin⁡xf(x) = \sin x and f(x)=cos⁡xf(x) = \cos x, the domain is all real numbers R\mathbb{R}, and the range is restricted to [−1,1][-1, 1]. For f(x)=tan⁡xf(x) = \tan x, the domain excludes values where cos⁡x=0\cos x = 0 (i.e., x=(2n+1)π2x = (2n+1)\frac{\pi}{2}), resulting in a range of (−∞,∞)(-\infty, \infty).

•

Graphs of Sine and Cosine: The graph of y=sin⁡xy = \sin x is a continuous wave starting at the origin (0,0)(0,0), reaching a peak of 11 at π2\frac{\pi}{2}, and crossing the x-axis at π\pi. The graph of y=cos⁡xy = \cos x starts at its maximum value (0,1)(0,1) and crosses the x-axis at π2\frac{\pi}{2}. Both graphs oscillate smoothly between 11 and −1-1.

•

Periodicity: Trigonometric functions are periodic, meaning they repeat their values after a specific interval. The period of sin⁡x\sin x, cos⁡x\cos x, sec⁡x\sec x, and csc⁡x\csc x is 2π2\pi. The period of tan⁡x\tan x and cot⁡x\cot x is π\pi. Visually, this means the entire shape of the graph repeats horizontally every 2π2\pi (or π\pi) units.

•

Graph Transformations: For a function y=asin⁡(bx+c)+dy = a \sin(bx + c) + d, 'aa' determines the amplitude (the height from the center line to the peak). 'bb' determines the period through the formula P=2π∣b∣P = \frac{2\pi}{|b|}. 'cc' shifts the graph horizontally (phase shift), and 'dd' shifts the graph vertically (midline).

📐Formulae

π radians=180∘\pi \text{ radians} = 180^{\circ}

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

1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta

1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B

cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

sin⁡2θ=2sin⁡θcos⁡θ=2tan⁡θ1+tan⁡2θ\sin 2\theta = 2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}

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

Period (T)=2π∣b∣ for sin⁡(bx) or cos⁡(bx)\text{Period } (T) = \frac{2\pi}{|b|} \text{ for } \sin(bx) \text{ or } \cos(bx)

💡Examples

Problem 1:

Given cos⁡x=−35\cos x = -\frac{3}{5} and xx lies in the third quadrant (π<x<3π2\pi < x < \frac{3\pi}{2}), find the values of sin⁡x\sin x and tan⁡x\tan x.

Solution:

Step 1: Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. sin⁡2x+(−35)2=1\sin^2 x + (-\frac{3}{5})^2 = 1 sin⁡2x+925=1\sin^2 x + \frac{9}{25} = 1 sin⁡2x=1−925=1625\sin^2 x = 1 - \frac{9}{25} = \frac{16}{25} sin⁡x=±1625=±45\sin x = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} Step 2: Determine the sign based on the quadrant. In the third quadrant, sin⁡x\sin x is negative. So, sin⁡x=−45\sin x = -\frac{4}{5}. Step 3: Calculate tan⁡x\tan x using tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}. tan⁡x=−4/5−3/5=43\tan x = \frac{-4/5}{-3/5} = \frac{4}{3}

Explanation:

We use the fundamental Pythagorean identity to find the magnitude of the missing function and then apply the ASTC rule to determine the correct sign based on the specified quadrant.

Problem 2:

Determine the amplitude and period of the function y=4sin⁡(3x)y = 4 \sin(3x) and describe its graph compared to y=sin⁡xy = \sin x.

Solution:

Step 1: Identify the amplitude 'aa'. In y=asin⁡(bx)y = a \sin(bx), a=4a = 4. Amplitude =∣4∣=4= |4| = 4. Step 2: Identify the frequency coefficient 'bb'. Here b=3b = 3. Step 3: Calculate the period using P=2πbP = \frac{2\pi}{b}. P=2π3P = \frac{2\pi}{3} Step 4: Comparison. The graph of y=4sin⁡(3x)y = 4 \sin(3x) is vertically stretched by a factor of 4 (reaching peaks at 44 and troughs at −4-4) and horizontally compressed by a factor of 3 (completing one full cycle every 2π3\frac{2\pi}{3} instead of 2π2\pi).

Explanation:

The amplitude is the absolute value of the leading coefficient, representing the peak height. The period is calculated by dividing the standard period (2π2\pi) by the coefficient of xx.