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Trigonometry - Trigonometric Functions and their Graphs

Grade 11ICSE

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

🔑Concepts

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The Sine function y=sin⁡xy = \sin x is a periodic function with a period of 2π2\pi. It oscillates between a maximum value of 11 and a minimum value of −1-1. The graph passes through the origin (0,0)(0,0) and has xx-intercepts at nπn\pi, where nn is an integer.

Graph of the sine function showing periodicity and amplitude.
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The Cosine function y=cos⁡xy = \cos x also has a period of 2π2\pi and an amplitude of 11. Unlike the sine function, it is an even function (symmetric about the yy-axis) and starts at its maximum value (0,1)(0,1) when x=0x=0.

Graph of the cosine function.
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The Tangent function y=tan⁡xy = \tan x has a period of π\pi. It is undefined at x=(2n+1)π2x = (2n+1)\frac{\pi}{2} (where cos⁡x=0\cos x = 0), resulting in vertical asymptotes at these points. The range of the tangent function is (−∞,∞)(-\infty, \infty).

Graph of the tangent function with vertical asymptotes.
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Signs of Trigonometric Functions: In the Cartesian plane, the signs of trig functions depend on the quadrant: All are positive in Quadrant I, Sine in Quadrant II, Tangent in Quadrant III, and Cosine in Quadrant IV (CAST rule).

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Transformations: For y=asin⁡(bx+c)+dy = a \sin(bx + c) + d, ∣a∣|a| is the amplitude, 2π∣b∣\frac{2\pi}{|b|} is the period, −cb-\frac{c}{b} is the phase shift (horizontal), and dd is the vertical shift.

📐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.

Problem 3:

Sketch the graph of f(x)=2sin⁡(2x)f(x) = 2\sin(2x) for the interval 0≤x≤π0 \le x \le \pi and identify its amplitude and period.

Graph of y = 2sin(2x) from 0 to pi.

Solution:

  1. Identify parameters: For f(x)=2sin⁡(2x)f(x) = 2\sin(2x), we have a=2a = 2 and b=2b = 2.
  2. Amplitude: The amplitude is ∣a∣=2|a| = 2.
  3. Period: The period is T=2πb=2π2=πT = \frac{2\pi}{b} = \frac{2\pi}{2} = \pi.
  4. Key Points: Since the period is π\pi, we divide it into four quarters: 0,π4,π2,3π4,π0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi. At x=0x=0, y=0y=0. At x=π4x=\frac{\pi}{4}, y=2sin⁡(π2)=2y=2\sin(\frac{\pi}{2}) = 2. At x=π2x=\frac{\pi}{2}, y=2sin⁡(π)=0y=2\sin(\pi) = 0. At x=3π4x=\frac{3\pi}{4}, y=2sin⁡(3π2)=−2y=2\sin(\frac{3\pi}{2}) = -2. At x=πx=\pi, y=2sin⁡(2π)=0y=2\sin(2\pi) = 0.

Explanation:

The coefficient 2 outside the sine function stretches the graph vertically (amplitude), while the coefficient 2 inside the sine function compresses the graph horizontally, halving the period from 2π2\pi to π\pi.

Problem 4:

Find the values of cos⁡x\cos x and tan⁡x\tan x if sin⁡x=513\sin x = \frac{5}{13} and xx is in the second quadrant (90∘<x<180∘90^{\circ} < x < 180^{\circ}).

Reference triangle in the second quadrant for sin x = 5/13.

Solution:

  1. Using Identity: We know sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. (513)2+cos⁡2x=1\left(\frac{5}{13}\right)^2 + \cos^2 x = 1 25169+cos⁡2x=1\frac{25}{169} + \cos^2 x = 1 cos⁡2x=1−25169=144169\cos^2 x = 1 - \frac{25}{169} = \frac{144}{169}
  2. Determine Sign: In the second quadrant, cos⁡x\cos x is negative. cos⁡x=−144169=−1213\cos x = -\sqrt{\frac{144}{169}} = -\frac{12}{13}
  3. Find Tangent: tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} tan⁡x=5/13−12/13=−512\tan x = \frac{5/13}{-12/13} = -\frac{5}{12}

Explanation:

In Quadrant II, only Sine (and Cosecant) is positive. Both Cosine and Tangent must result in negative values. The Pythagorean triplet (5,12,13)(5, 12, 13) is used here.