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Trigonometric Functions - Trigonometric Functions and Signs

Grade 11CBSE

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

🔑Concepts

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Trigonometric functions are defined using a unit circle (a circle with radius r=1r=1 centered at the origin). For any angle xx, the point P(a,b)P(a, b) on the circle is represented as a=cos⁡xa = \cos x and b=sin⁡xb = \sin x. Visually, the x-coordinate of a point on the circumference tracks the cosine value, while the y-coordinate tracks the sine value as the point rotates.

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The signs of trigonometric functions are determined by the quadrant in which the terminal side of the angle lies. In Quadrant I (00 to π2\frac{\pi}{2}), all functions are positive. In Quadrant II (π2\frac{\pi}{2} to π\pi), only sin⁡\sin and csc⁡\csc are positive. In Quadrant III (π\pi to 3π2\frac{3\pi}{2}), only tan⁡\tan and cot⁡\cot are positive. In Quadrant IV (3π2\frac{3\pi}{2} to 2π2\pi), only cos⁡\cos and sec⁡\sec are positive. This is visually represented by the 'ASTC' rule (All Silver Tea Cups).

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Trigonometric functions are periodic, meaning their values repeat after a fixed interval. The functions sin⁡x\sin x and cos⁡x\cos x have a period of 2π2\pi, while tan⁡x\tan x has a period of π\pi. Visually, the graphs of sine and cosine are continuous waves that complete one full cycle every 2π2\pi units along the x-axis.

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The domain of sin⁡x\sin x and cos⁡x\cos x is the set of all real numbers RR. Their range is restricted to the interval [−1,1][-1, 1]. On a coordinate graph, this signifies that the curves for sine and cosine are trapped vertically between the lines y=1y=1 and y=−1y=-1.

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The function tan⁡x\tan x is defined as sin⁡xcos⁡x\frac{\sin x}{\cos x} and is undefined whenever cos⁡x=0\cos x = 0. Visually, this occurs at odd multiples of π2\frac{\pi}{2} (e.g., ±π2,±3π2\pm\frac{\pi}{2}, \pm\frac{3\pi}{2}), where the graph of tangent displays vertical asymptotes.

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Functions exhibit specific symmetries: cos⁡(−x)=cos⁡x\cos(-x) = \cos x is an even function, appearing visually symmetric about the y-axis. sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x and tan⁡(−x)=−tan⁡x\tan(-x) = -\tan x are odd functions, appearing visually symmetric about the origin (0,0)(0,0).

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Special values at quadrantal angles are essential for calculation. At 00, sin⁡\sin is 00 and cos⁡\cos is 11. At π2\frac{\pi}{2}, sin⁡\sin is 11 and cos⁡\cos is 00. At π\pi, sin⁡\sin is 00 and cos⁡\cos is −1-1. At 3π2\frac{3\pi}{2}, sin⁡\sin is −1-1 and cos⁡\cos is 00. These values correspond to the intersections of the unit circle with the x and y axes.

📐Formulae

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x

1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x

tan⁡x=sin⁡xcos⁡x,x≠(2n+1)π2\tan x = \frac{\sin x}{\cos x}, x \neq (2n+1)\frac{\pi}{2}

cot⁡x=cos⁡xsin⁡x,x≠nπ\cot x = \frac{\cos x}{\sin x}, x \neq n\pi

sin⁡(2nπ+x)=sin⁡x,n∈Z\sin(2n\pi + x) = \sin x, n \in Z

cos⁡(2nπ+x)=cos⁡x,n∈Z\cos(2n\pi + x) = \cos x, n \in Z

sin⁡x=0  ⟹  x=nπ,n∈Z\sin x = 0 \implies x = n\pi, n \in Z

cos⁡x=0  ⟹  x=(2n+1)π2,n∈Z\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}, n \in Z

💡Examples

Problem 1:

If cos⁡x=−35\cos x = -\frac{3}{5} and xx lies in the third quadrant, find the values of the other five trigonometric functions.

Solution:

  1. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1: sin⁡2x=1−(−35)2=1−925=1625\sin^2 x = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}.
  2. In the third quadrant, sin⁡x\sin x is negative. Therefore, sin⁡x=−1625=−45\sin x = -\sqrt{\frac{16}{25}} = -\frac{4}{5}.
  3. Calculate tan⁡x=sin⁡xcos⁡x=−4/5−3/5=43\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \frac{4}{3}.
  4. Calculate cot⁡x=1tan⁡x=34\cot x = \frac{1}{\tan x} = \frac{3}{4}.
  5. Calculate sec⁡x=1cos⁡x=−53\sec x = \frac{1}{\cos x} = -\frac{5}{3}.
  6. Calculate csc⁡x=1sin⁡x=−54\csc x = \frac{1}{\sin x} = -\frac{5}{4}.

Explanation:

The solution involves finding sin⁡x\sin x using the Pythagorean identity and then determining the correct sign based on the quadrant (Quadrant III: tan⁡\tan and cot⁡\cot are positive; others are negative). Once sin⁡\sin and cos⁡\cos are known, the reciprocal and quotient identities are used for the rest.

Problem 2:

Find the value of sin⁡31π3\sin \frac{31\pi}{3}.

Solution:

  1. Express the angle in terms of multiples of 2π2\pi: 31π3=(10π+π3)\frac{31\pi}{3} = (10\pi + \frac{\pi}{3}).
  2. Note that 10π10\pi is 5×2π5 \times 2\pi. Since the period of sin⁡\sin is 2π2\pi, sin⁡(2nπ+θ)=sin⁡θ\sin(2n\pi + \theta) = \sin \theta.
  3. Therefore, sin⁡(10π+π3)=sin⁡π3\sin(10\pi + \frac{\pi}{3}) = \sin \frac{\pi}{3}.
  4. The value of sin⁡π3\sin \frac{\pi}{3} is 32\frac{\sqrt{3}}{2}.

Explanation:

This approach uses the periodicity of trigonometric functions. By breaking down a large angle into a multiple of 2π2\pi plus a remainder, we can reduce the problem to finding the value of a standard acute angle.