Trigonometric Function: Definition, Formula, & Facts

Trigonometric functions are fundamental in mathematics, particularly in geometry, calculus, and applied mathematics. They are used to describe relationships involving lengths and angles in right triangles. The graph of trigonometric functions helps in finding the domain and its range. The six basic trigonometric functions are sine, cosine, tangent, cosecant, secant, and cotangent are the trigonometric functions.

In this article, we will cover the concepts of the trigonometric function. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including one in 2021, and one in 2023.

What are Trigonometric Function?

In trigonometry, there are six basic trigonometric functions. These functions are trigonometric ratios that are based on ratios of sides in a right triangle: the hypotenuse (the longest side), the base (the side adjacent to a chosen angle), and the perpendicular (the side opposite the chosen angle). These functions are sine, cosine, tangent, secant, cosecant, and cotangent. They help us find different values in triangles by comparing these side lengths.

Trigonometric Functions Formulas

Using the sides of a right-angled triangle, we may use specific formulas to get the values of the trigonometric functions. We utilize the shortened form of these functions to write these formulas. The notation for sine is sin; the notation for cosine is cos; the notation for tangent is tan; the notation for secant is sec; the notation for cosecant is cosec; and the notation for cotangent is cot. The basic formulas to find the trigonometric functions are as follows:

• sin θ = Perpendicular/Hypotenuse
• cos θ = Base/Hypotenuse
• tan θ = Perpendicular/Base
• sec θ = Hypotenuse/Base
• cosec θ = Hypotenuse/Perpendicular
• cot θ = Base/Perpendicular

As we can observe from the above-given formulas, sine and cosecant are reciprocals of each other. Similarly, the reciprocal pairs are cosine and secant, and tangent and cotangent.

Trigonometric Functions Values

The trigonometric functions' domain θ can be represented in either degrees or radians. A table showing some of the principal values of θ for the different trigonometric functions can be seen below. These principal values, usually referred to as standard values of the trig function at specific angles, are frequently used in computations. The principal values of trigonometric functions have been found from a unit circle. Additionally, these numbers satisfy all trigonometric formulas.

Domain, range, and graph of trigonometric functions

Sine Function

y = f(x) = sin(x)

Domain is R

Range is [-1, 1]

Cosine Function

y = f(x)= cos(x)

Domain is R

Range is [-1, 1]

Tangent Function

y = f(x) = tan(x)

Domain is $\mathbb{R}-\left\{\frac{(2 \mathrm{n}+1) \pi}{2}, \mathrm{n} \in \mathbb{I}\right\}$

Range is R

Cosecant Function

y = f(x) = cosec(x)

Domain is R - {nπ, n ∈ I (Integers)}

Range is R - (1, 1)

Secant Function

y = f(x) = sec(x)

Domain is $\mathbb{R}-\left\{\frac{(2 \mathrm{n}+1) \pi}{2}, \mathrm{n} \in \mathbb{I}\right\}$

Range is R - (-1, 1)

Cotangent Function

y = f(x) = cot(x)

Domain is R - {nπ, n ∈ I (Integers)}

Range is R

Summary

In order to solve problems involving angles and distance, trigonometric functions are crucial. They are essential instruments in many scientific and engineering fields because of their qualities and uses, which go well beyond perfect triangles. A crucial component of the trigonometric function analysis is understanding its domain and range.

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Solved Examples Based On the Trigonometric Functions:

Example 1: Let $f(x)=\sin ^{-1} x$ and $g(x)=\frac{x^2-x-2}{2 x^2-x-6}$. If $g(2)=\lim _{x \rightarrow 2} g(x)$, then the domain of the function fog is:
1) $(-\infty,-1] \cup[2, \infty)$
2) $(-\infty,-2] \cup[-1, \infty)$
3) $(-\infty,-2] \cup\left[-\frac{3}{2}, \infty\right)$
4) $(-\infty,-2] \cup\left[-\frac{4}{3}, \infty\right)$Solution:

$\begin{aligned}
& \text { Domain of } f \circ g(\mathrm{x})=\sin ^{-1}(\mathrm{~g}(\mathrm{x})) \\
& \Rightarrow \lg (\mathrm{x})!\leq 1 \\
& \text { as }-1 \leq \sin ^{-1} x \leq 1 \\
& g(2)=\lim _{x \rightarrow 2} g(x) \\
& g(2)=\lim _{x \rightarrow 2} \frac{x^2-x-2}{2 x^2-x-6}=\lim _{x \rightarrow 2} \frac{2 x-1}{4 x-1}=\frac{3}{7} \\
& \left|\frac{x^2-x-2}{2 x^2-x-6}\right| \leq 1 \\
& \left.\frac{(x+1)(x-2)}{(2 x+3)(x-2)} \right\rvert\, \leq 1 \\
& \frac{x+1}{2 x+3} \leq 1 \text { and } \frac{x+1}{2 x+3} \geq-1 \\
& \frac{x+1-2 x-3}{2 x+3} \leq 0 \text { and } \frac{x+1+2 x+3}{2 x+3} \geq 0 \\
& \frac{x+2}{2 x+3} \geq 0 \text { and } \frac{3 x+4}{2 x+3} \geq 0 \\
& x \in(-\infty,-2] \cup\left[-\frac{4}{3}, \infty\right)
\end{aligned}$
Hence, the answer is the option 4.

Example 2: What is the range of function $f(x)=2 \sin (x)$ ?
1) $[-2,2]$
2) $(-2,2)$
3) $[-2,2]-\{0\}$
4) $\mathbb{R}$

Solution:

As we have learned in

Sine Function: Range is [-1, 1]

Now,

Now,
The range of the basic $\sin$ function is $-1 \leq \sin (x) \leq 1$
Multiply the edges of the range by : 2
$-2 \leq 2 \sin (x) \leq 2$
Therefore the range is

$-2 \leq f(x) \leq 2$

Hence, the answer is the option 1 .
Example 3: Let $f_k(x)=\frac{1}{k}\left(\sin ^k x+\cos ^k x\right)$ where $x \in R$ and $k \geqslant 1 \quad$ Then $f_4(x)-f_6(x)$ equals :
Solution:

$\begin{aligned}
& \frac{1}{4}\left(\sin ^4 x+\cos ^4 x\right)-\frac{1}{6}\left(\sin ^6 x+\cos ^6 x\right) \\
= & \frac{1}{4}\left(1-2 \sin ^2 x \cos ^2 x\right)-\frac{1}{6}\left(\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x+\cos ^4 x-\sin ^4 x \cos ^4 x\right)\right) \\
= & \frac{1}{4}\left(1-2 \sin ^2 x \cos ^2 x\right)-\frac{1}{6}\left(1-2 \sin ^2 x \cos ^2 x-\sin ^2 x \cos ^2 x\right) \\
= & \frac{1}{4}\left(1-2 \sin ^2 x \cos ^2 x\right)-\frac{1}{6}\left(1-3 \sin ^2 x \cos ^2 x\right)=\frac{1}{12}
\end{aligned}$
Hence, the answer is 1/12.

Example 4: Let $f: \mathbb{R} \rightarrow \mathbb{R}_{\text {be a function defined by }} f(x)=\log _{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is
Solution:

$\begin{aligned}
\because & -\sqrt{2} \leq \sin x-\cos x \leq \sqrt{2} \\
\Rightarrow & -2 \leq \sqrt{2}(\sin x-\cos x) \leq 2 \\
\Rightarrow & m-4 \leq \sqrt{2}(\sin x-\cos x)+m-2 \leq m \\
\Rightarrow & \log _{\sqrt{m}}^{(m-4)} \leq \log _{\sqrt{m}}^{(\sqrt{2}(\sin x-\cos x)+m-2\}} \leq \log _{\sqrt{m}}^{\mathrm{m}} \\
& \Downarrow \\
& 0 \\
\Rightarrow & \log _{\sqrt{m}}^{(\mathrm{m}-4)}=0 \\
\Rightarrow & \mathrm{m}=5
\end{aligned}$

Hence, the answer is 5.