Of the Six Basic Trigonometric Functions Which Are Continuous Over All Real Numbers
Basics of Trigonometry: Value Table, Identities
Basics of trigonometry: Mathematics is divided into several branches, each with its significance. Trigonometry is a vital branch of Mathematics that investigates the relationship between angles of a right-angled triangle and the lengths of its sides. Trigonometry can be used to compute the heights of mountains in Geology; it can also be used to calculate the distance between stars or planets in Astronomy. It is employed in Physics and Architecture. We are familiar with trigonometric ratios for acute angles as the ratio of sides of a right-angled triangle. Let us now apply the definition to other types of angles measured in radians and study it as a trigonometric function.
Trigonometric Functions
Consider a unit circle with its centre at the origin of the coordinate axes. Let \(P(a, b)\) be any point on the circle with angle \(A O P=x\) radian, i.e., length of \(\operatorname{arc} A P=x\).
We define cosine and sine functions of radian measure \(x\) as follows :
\(\cos x=a\) and \(\sin x=b\)
\(\Rightarrow \cos x=x-\) coordinate of the point \(P\) on the unit circle
And, \(\sin x=y-\) coordinate of point \(P\).
Remark 1: \(x\) is the length of \(\operatorname{arc} A P\) of the unit circle. Therefore, \(\cos x\) and \(\sin x\) are also known as circular functions of the real variable \(x\).
Remark 2: \(\triangle O M P\) is a right triangle that is right-angled at \(M\). The trigonometric ratios \(\angle M O P\) are \(\cos \angle M O P=\frac{O M}{O P}=\frac{a}{1}=a\) and \(\sin \angle M O P=\frac{P M}{O P}=\frac{b}{1}=b\)
\(\Rightarrow \cos \angle A O P=a\) and \(\sin \angle A O P=b\)
\(\Rightarrow \cos \angle A O P=\cos x\) and \(\sin \angle A O P=\sin x\)
Thus, the trigonometric ratios sine and cosine of an acute angle of radian measure \(x\) are the same as the corresponding trigonometric function of a real number \(x\).
Remark 3: From the above definition, if \(P\) is a point on the unit circle such that length of \(\operatorname{arc}(A P)=x\), then the coordinates of the point \(P\) are \((\cos x, \sin x)\).
Sine and Cosine Functions
Consider a unit circle with its centre at the origin. Suppose the circle cuts the coordinate axes at \(A, B, C\) and \(D\). The coordinates of these points are \(A(1,0), B(0,1), C(-1,0)\), and \(D(0,-1)\).
Since one complete revolution subtends an angle of \(2 \pi\) radian at the centre of the circle, \(\angle A O B=\) \(\frac{\pi}{2}, \angle A O C=\pi\) and \(\angle A O D=\frac{3 \pi}{2}\).
Let us now find the values of sine and cosine functions at \(0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}\) and \(2 \pi\).
We have seen that if \(P(a, b)\) be any point on the circle with angle \(A O P=x\) radian i.e., length of \(\operatorname{arc} A P=x\) then, \(\cos x=a\), and \(\sin x=b\).
Angle \(x\) | New coordinates of point \(P\) at \(x\) | Value of Sine and Cosine of \(x\) |
\(x=0\) | \((1,0)\) | \(\cos 0=1, \sin 0=0\) |
\(x=\frac{\pi}{2}\) | \((0,1)\) | \(\cos \frac{\pi}{2}=0, \sin \frac{\pi}{2}=1\) |
\(x=\pi\) | \((-1,0)\) | \(\cos \pi=-1, \sin \pi=0\) |
\(x=\frac{3 \pi}{2}\) | \((0,-1)\) | \(\cos \frac{3 \pi}{2}=0, \sin \frac{3 \pi}{2}=-1\) |
\(x=2 \pi\) | \((1,0)\) | \(\cos 2 \pi=1, \sin 2 \pi=0\) |
We can now take one complete revolution from point \(P\). As a result, we can see that the values of the sine and cosine functions do not change as \(x\) increases (or decreases) by an integral multiple of \(2 \pi\). Thus,
\(\sin (2 n \pi+x)=\sin x, n \in \mathrm{Z}, \cos (2 n \pi+x)=\cos x, n \in Z\)
It is evident from the above figure that:
- \(\sin x=0\), if \(x=0, \pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots\), i.e. when \(x\) is an integral multiple of \(\pi\).
- \(\cos x=0\), if \(x=\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \ldots\), i.e. \(\cos x\) vanishes when \(x\) is an odd multiple of \(\frac{\pi}{2}\).
Thus, \(\sin x=0\) implies \(x=n \pi\), where \(n\) is any integer.
\(\cos x=0\) implies \(x=(2 n+1) \frac{\pi}{2}\), where \(n\) is any integer.
Other Trigonometric Functions
We now define other trigonometric functions in terms of sine and cosine functions.
\(\operatorname{cosec} x=\frac{1}{\sin x}, x \neq n \pi\), where \(n\) is any integer.
\(\sec x=\frac{1}{\cos x}, x \neq(2 n+1) \frac{\pi}{2}\), where \(n\) is any integer.
\(\tan x=\frac{\sin x}{\cos x}, x \neq(2 n+1) \frac{\pi}{2}\), where \(n\) is any integer
\(\cot x=\frac{\cos x}{\sin x}, x \neq n \pi\), where \(n\) is any integer.
The values of trigonometric functions for some basic angles are as shown below.
Trigonometric Identities
- \(\cos ^{2} x+\sin ^{2} x=1\) for all \(x \in R\)
- \(1+\tan ^{2} x=\sec ^{2} x\) for all \(x \neq(2 n-1) \frac{\pi}{2}\)
- \(1+\cot ^{2} x=\operatorname{cosec}^{2} x\) for all \(x \in R – \left\{ {n\pi :n \in Z} \right\}\)
Proofs of Trigonometric Identities
Consider a unit circle with its centre at the origin. Let \(P(a, b)\) be a point on the circle such that \(\widehat{A P}=x\). Then, \(\angle A O P=x\). Using the definition of trigonometric functions \(\cos x\) and \(\sin x\), we obtain
\(a=\cos x\)
\(b=\sin x\)
i. \(\cos ^{2} x+\sin ^{2} x=1\) for all \(x \in R\)
Proof: Now, \(O P=1\)
\(\Rightarrow \sqrt{(a-0)^{2}+(b-0)^{2}}=1\)
\(\Rightarrow a^{2}+b^{2}=1\)
\(\therefore \cos ^{2} x+\sin ^{2} x=1\)
ii. \(1+\tan ^{2} x=\sec ^{2} x\) for all \(x \neq(2 n-1) \frac{\pi}{2}\)
Proof: We know that \(\sin{ }^{2} x+\cos ^{2} x=1\)
If \(x \neq(2 n-1) \frac{\pi}{2}\), then \(\cos x \neq 0\). So, dividing throughout by \(\cos ^{2} x\), we obtain
\(\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x}=\frac{1}{\cos ^{2} x}\) for all \(x \neq(2 n-1) \frac{\pi}{2}\)
\(\Rightarrow \frac{\cos ^{2} x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}=\frac{1}{\cos ^{2} x}\) for all \(x \neq(2 n-1) \frac{\pi}{2}\)
\(\Rightarrow 1+\tan ^{2} x=\sec ^{2} x\) for all \(x \neq(2 n-1) \frac{\pi}{2}\).
iii. \(1+\cot ^{2} x=\operatorname{cosec}^{2} x\) for all \(x \in R – \left\{ {n\pi :n \in Z} \right\}\)
Proof: We know that \(\cos ^{2} x+\sin ^{2} x=1\)
If \(x \neq n \pi\), then \(\sin x \neq 0\). So, dividing both sides by \(\sin^{2} x\), we obtain
\(\frac{\cos ^{2} x+\sin ^{2} x}{\sin ^{2} x}=\frac{1}{\sin ^{2} x}\) for all \(x \neq n \pi\)
\(\Rightarrow \frac{\cos ^{2} x}{\sin ^{2} x}+\frac{\sin ^{2} x}{\sin ^{2} x}=\frac{1}{\sin ^{2} x}\) for all \(x \neq n \pi\)
\(\Rightarrow \cot ^{2} x+1=\operatorname{cosec}^{2} x\) for all \(x \neq n \pi\)
\(\Rightarrow 1+\cot ^{2} x=\operatorname{cosec}^{2} x\) for all \(x \neq n \pi\)
These identities are also known as the Pythagorean Identities.
Signs of Trigonometric Functions
Let \(O\) be the centre of a unit circle and \(A\) be the point \((1,0)\). Let \(P(a, b)\) be a point on the unit circle such that length of \(\widehat{A P}=x\), then \(\angle A O P=x\),
Thus, the six trigonometric functions are defined as:
- \(\sin x=b\),for all \(x \in R\)
- \(\cos x=a\), for all \(x \in R\)
- \(\tan x=\frac{b}{a}, x \neq(2 n+1) \frac{\pi}{2}\), where \(n\) is any integer.
- \(\cot x=\frac{a}{b}, x \neq n \pi, n\) is any integer.
- \(\sec x=\frac{1}{a}, x \neq(2 n+1) \frac{\pi}{2}, n\) is any integer.
- \(\operatorname{cosec} x=\frac{1}{b}, x \neq n \pi, n\) is any integer.
Note that in the unit circle, \(-1 \leq a \leq 1\) and \(-1 \leq b \leq 1\)
Also, we know that
- \(a>0, b>0\) in I quadrant
- \(a<0, b>0\) in II quadrant
- \(a<0, b<0\) in III quadrant
- \(a>0, b<0\) in IV quadrant
Therefore, the sign of trigonometric functions in various quadrants is shown below.
This can be summarised as shown here.
Quadrant | I | II | III | IV |
Trigonometric functions which are \(+ve\) | All | \(\sin x\) | \(\tan x\) | \(\cos x\) |
\(\operatorname{cosec} x\) | \(\cot x\) | \(\sec x\) |
Domain of Trigonometric Functions
Function | Domain |
\(\sin,\cos\) | all real numbers |
\(\tan,\sec\) | all real numbers other than \((2 n+1) \frac{\pi}{2}, n \in Z\) |
\(\cot , \operatorname{cosec}\) | all real numbers other than \(n \pi, n \in Z\) |
Range of Trigonometric Functions
As \(-1 \leq a \leq 1\) and \(-1 \leq b \leq 1\) in a unit circle.
\(\Rightarrow-1 \leq \cos x \leq 1\) and \(-1 \leq \sin x \leq 1\)
Thus, the maximum and minimum values of \(\sin x\) and \(\cos x\) are \(1\) and \(-1\), respectively.
Since \(\tan x=\frac{b}{a}\) and \(\cot x=\frac{a}{b}\), and any of \(a\) and \(b\) can be greater than the other, \(\tan x\) and \(\cot x\) can take any real value.
Now, \(-1 \leq a \leq 1, a \neq 0 \Rightarrow \frac{1}{a} \geq 1\) or \(\frac{1}{a} \leq-1\)
\(\Rightarrow \sec x \geq 1\) or \(\sec x \leq 1\)
Also, \(-1 \leq b \leq 1, b \neq 0 \Rightarrow \frac{1}{b} \geq 1\) or \(\frac{1}{b} \leq-1\)
\(\Rightarrow \operatorname{cosec} x \geq 1\) or \(\operatorname{cosec} x \leq-1\)
Thus, we have
Function | Range |
\(\sin,\cos\) | \([-1,1]\) |
\(\tan,\cot\) | any real value |
\(\sec, \operatorname{cosec}\) | any real value except \((-1,1)\) |
Behaviour of Trigonometric Functions
Consider a unit circle centred at the origin \(O\) of the coordinate axes. The circle cuts the coordinates axes at \(A(1,0), B(0,1), C(-1,0)\) and \(D(0,-1)\). Let \(P(a, b)\) be a point on the circle whose equation is \(x^{2}+y^{2}=1\) such that \(\operatorname{arc} A P=x\) or equivalently radian measure of \(\angle A O P\) is \(x\).
Then, \(a=\cos x\) and \(b=\sin x\)
We observe that in the first quadrant as \(x\) increases from \(0\) to \(\frac{\pi}{2}, b\) increases from \(0\) to \(1\) , and \(a\) decreases from \(1\) to \(0\). Since \(b=\sin x\), and \(a=\cos x\). Thus, in the first quadrant \(\sin x\) increases from \(0\) to \(1\), and \(\cos x\) decreases from \(1\) to \(0\). In the second quadrant as \(x\) increases from \(\frac{\pi}{2}\) to \(\pi, b\) decreases from \(1\) to \(0\), and \(a\) decreases from \(0\) to \(-1\). Thus, in the second quadrant \(\cos x\) decreases from \(0\) to \(-1\), and \(\sin x\) decreases from \(1\) to \(0\). Similarly, as \(x\) increases from \(\pi\) to \(\frac{3 \pi}{2}, b\) decreases from \(0\) to \(-1\), and \(a\) increases from \(-1\) to \(0\). Thus, in the third quadrant \(\cos x\) increases from \(-1\) to \(0\), and \(\sin x\) decreases from \(0\) to \(-1\). Now, as \(x\) increases from \(\frac{3 \pi}{2}\) to \(2 \pi, b\) increases from \(-1\) to \(0\), and \(a\) increases from \(0\) to \(1\).
Similarly, we can observe the variations in the values of other trigonometric functions. The following image exhibits the same.
Solved Examples – Basics of Trigonometry
Q.1. Find \(\sin x\) and \(\tan x\), if \(\cos x=-\frac{12}{13}\) and \(x\) lies in the third quadrant.
Ans: We know that
\(\cos ^{2} x+\sin ^{2} x=1\)
\(\Rightarrow \sin x=\pm \sqrt{1-\cos ^{2} x}\)
Since in the III quadrant, \(\sin x\) is negative, we get
\(\therefore \sin x=-\sqrt{1-\cos ^{2} x}\)
\(\Rightarrow \sin x=-\sqrt{1-\left(-\frac{12}{13}\right)^{2}}=-\frac{5}{13}\)
And \(\tan x=\frac{\sin x}{\cos x}\)
\(\Rightarrow \tan x=-\frac{5}{13} \times \frac{13}{-12}=\frac{5}{12}\)
Hence, \(\sin x=-\frac{5}{13}\) and \(\tan x=\frac{5}{12}\)
Q.2. If \(10 \sin ^{4} \alpha+15 \cos ^{4} \alpha=6\), find the value of \(27 \operatorname{cosec}^{6} \alpha+8 \sec ^{6} \alpha\)
Ans: Consider
\(10 \sin ^{4} \alpha+15 \cos ^{4} \alpha=6\)
\(\Rightarrow 10 \sin ^{4} \alpha+15 \cos ^{4} \alpha=6\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}\)
Dividing both sides by \(\cos ^{4} \alpha\)
\(\Rightarrow 10 \tan ^{4} \alpha+15=6\left(\tan ^{2} \alpha+1\right)^{2}\)
\(\Rightarrow\left(2 \tan ^{2} \alpha-3\right)^{2}=0\)
\(\Rightarrow \tan ^{2} \alpha=\frac{3}{2}\)
\(\therefore 27 \operatorname{cosec}^{6} \alpha+8 \sec ^{6} \alpha=27\left(1+\cot ^{2} \alpha\right)^{3}+8\left(1+\tan ^{2} \alpha\right)^{3}\)
\(=27\left(1+\frac{2}{3}\right)^{3}+8\left(1+\frac{3}{2}\right)^{3}\)
\(=27 \times \frac{125}{27}+8 \times \frac{125}{8}\)
\(\therefore 27 \operatorname{cosec}^{6} \alpha+8 \sec ^{6} \alpha=250\)
Q.3. If \(\tan x=-\frac{5}{12}\) and \(x\) lies in the second quadrant, find the values of five other trigonometric functions.
Ans: Given: \(\tan x=-\frac{5}{12}\) and \(x\) lies in the III quadrant.
\(\therefore \cot x=\frac{1}{\tan x}=-\frac{12}{5}\)
We know that \(\sec ^{2} x=1+\tan ^{2} x\)
\(\Rightarrow \sec ^{2} x=1+\left(-\frac{5}{12}\right)^{2}\)
\(=1+\frac{25}{144}\)
\(=\frac{169}{144}\)
\(\Rightarrow \sec x=\pm \frac{13}{12}\)
But \(x\) lies in the second quadrant and \(\sec x\) is \(-ve\) in the second quadrant, therefore,
\(\sec x=-\frac{13}{12}\)
\(\therefore \cos x=\frac{1}{\sec x}=-\frac{12}{13}\)
Further, \(\sin x=\frac{\sin x}{\cos x} \cdot \cos x\)
\(=\tan x \cos x\)
\(=\left(-\frac{5}{12}\right) \times\left(-\frac{12}{13}\right)\)
\(\sin x=\frac{5}{13}\)
\(\therefore \operatorname{cosec} x=\frac{1}{\sin x}=\frac{13}{5}\)
Q.4. Find the value of \(\sin \left(-\frac{11 \pi}{3}\right)\).
Ans: Given: \(\sin \left(-\frac{11 \pi}{3}\right)\)
\(=\sin \left(-4 \pi+\frac{\pi}{3}\right)\)
\(=\sin \left((-2) 2 \pi+\frac{\pi}{3}\right)\)
\(=\sin \frac{\pi}{3}(\because \sin (2 n \pi+x)=\sin x)\)
\(\therefore \sin \left(-\frac{11 \pi}{3}\right)=\frac{\sqrt{3}}{2}\)
Q.5. Find the value of the following \(\operatorname{cosec}\left(-1710^{\circ}\right)\)
Ans: \(\operatorname{cosec}\left(-1410^{\circ}\right)=\operatorname{cosec}\left(-4 \times 360^{\circ}+30^{\circ}\right)\)
\(=\operatorname{cosec}\left((-4) 2 \pi+\frac{\pi}{6}\right)\)
\(=\operatorname{cosec} \frac{\pi}{6}(\because \operatorname{cosec}(2 n \pi+x)=\operatorname{cosec} x)\)
\(\therefore \operatorname{cosec}\left(-1710^{\circ}\right)=2\)
Summary
Let \(P(a, b)\) be any point on the circle with \(\angle A O P=x\) radian, i.e., length of \(\widehat{A P}=x\). Then, sine and \(\operatorname{cosine~functions~are~defined~as~} \sin x=b\) and \(\cos x=a\). The other trigonometric functions can be written in terms of sine and cosine functions. The values of trigonometric functions at \(x=0, \frac{\pi}{2}, \frac{3 \pi}{2}\), and \(2 \pi\) is also provided. After an interval of \(2 \pi\), trigonometric functions repeat their values. The Pythagorean Identities are \(\sin ^{2} x+\cos ^{2} x=1,1+\tan ^{2} x=\sec ^{2} x\), and \(1+\cot ^{2} x=\operatorname{cosec}^{2} x\). Also, if \(x\) lies in the first quadrant then the sign of all trigonometric functions is positive, If \(x\) lies in the second quadrant then \(\sin x\) and \(\operatorname{cosec} x\) is positive, in the third quadrant: \(\tan x\), and \(\cot x\) are positive. And, in the fourth quadrant \(\cos x\) and \(\sec x\) are positive.
Frequently Asked Questions (FAQs)
Q.1. What are the 6 basic trigonometric functions?
Ans: The six basic trigonometric functions are
(i) Sine
(ii) Cosine
(iii) Tangent
(iv) Cotangent
(v) Secant
(vi) Cosecant
Q.2. What is the formula for the six basic trigonomteric functions?
Ans: Consider a right-angled triangle \(ABC\) which is given below
Then,
\(\sin \angle A=\frac{\text { perpendicular }}{\text { hypotenuse }}\)
\(\cos \angle A=\frac{\text { adjacent }}{\text { hypotenuse }}\)
\(\tan \angle A=\frac{\text { perpendicular }}{\text { adjacent }}\)
\(\sin A=\frac{1}{\operatorname{cosec} A}\)
\(\cos A=\frac{1}{\sec A}\)
\(\tan A=\frac{1}{\cot A}\)
Q.3.What are the applications of trigonometry in real life?
Ans: Trigonometry can be used to compute the heights of mountains; it can also be used to calculate the distance between stars or planets in Astronomy, and it is widely employed in Physics, Architecture, and in \(GPS\) navigation systems.
Q.4. What are trigonometric ratios used for?
Ans: If you know the lengths of two sides of a right triangle, you can use trigonometric ratios to compute the measurements of one (or both) of the acute angles.
Q.5. What are the \(3\) Pythagorean identities?
Ans: The \(3\) Pythagorean identities of trigonometric functions are as follows:
\(\cos ^{2} x+\sin ^{2} x=1\) for all \(x \in R\)
\(1+\tan ^{2} x=\sec ^{2} x\) for all \(x \neq(2 n-1) \frac{\pi}{2}\)
\(1+\cot ^{2} x=\operatorname{cosec}^{2} x\) for all \(x \in R – \left\{ {n\pi :n \in Z} \right\}\)
Study Trigonometry Table Here
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