INTRODUCTION TO TRIGONOMETRY-Notes
Maths - Notes
BASICS OF RIGHT-ANGLED TRIANGLES
Given such a triangle:
- The side opposite the right angle is called the hypotenuse—the longest side.
- The side opposite the angle under consideration is the perpendicular (or opposite side).
- The side adjacent to that angle (other than hypotenuse) is the base (or adjacent side).
The famous Pythagoras theorem underlies trigonometry:
\[(Hypotenuse)^2=(Base)^2+(Perpendicular)^2\] This relationship ensures that all trigonometric ratios are well-defined and interconnected.
TRIGONOMETRIC RATIOS OF AN ANGLE
For an acute angle \(\theta\) in a right triangle, the six trigonometric ratios are defined as:
- \(\bf sin\ \theta\):
\[sin\ \theta=\frac{Perpendicular}{Hypotenuse}\] - \(\bf cos\ \theta\):
\[cos\ \theta=\frac{Base}{Hypotenuse}\] - \(\bf tan\ \theta\):
\[cos\ \theta=\frac{Perpendicular}{Base}\] - \(\bf Cosecant(cosec\ \theta)\):
\[cosec\ \theta=\frac{1}{sin\ \theta}=\frac{Hypotenuse}{Perpendicular}\] - \(\bf Secant(sec\ \theta)\):
\[sec\ \theta=\frac{1}{cos\ \theta}=\frac{Hypotenuse}{Base}\] - \(\bf Cotangent(cot\ \theta)\):
\[cot\ \theta=\frac{1}{tan\ \theta}=\frac{Base}{Perpendicular}\]
Rhyme to remember these trigonometric ratios
\(P\)andit \(B\)adri \(P\)rasad
\(H\)ar \(H\)ar \(B\)ole
\[ \color{white} \begin{array}{|c|c|c|} \hline \sin\theta & \cos\theta & \tan\theta \\ \hline \color{red}P & \color{red}B & \color{red}P \\ \hline \color{red}H & \color{red}H & \color{red}B \\ \hline \text{cosec}\ \theta & \sec\theta & \cot\theta \\ \hline \end{array} \]
Explanation using the mnemonic:
In this trick, the highlighted letters \(P, B, P, H, H, B\) from the line “Pandit Badri Prasad Har Har Bole” are mapped to the trigonometric ratios \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\).
The first red row \((P, B, P)\) gives the first letters of the numerators: for \(\sin\theta\) and \(\tan\theta\) use \(P\)
(perpendicular / opposite side), and for \(\cos\theta\) use \(B\) (base / adjacent side).
Thus,
\(\sin\theta \Rightarrow P\), \(\cos\theta \Rightarrow B\), \(\tan\theta \Rightarrow P\).
The second red row \((H, H, B)\) gives the first letters of the denominators: for \(\sin\theta\) and \(\cos\theta\) use \(H\) (hypotenuse), and for \(\tan\theta\) use \(B\) (base). So you obtain \[ \sin\theta = \frac{P}{H}, \quad \cos\theta = \frac{B}{H}, \quad \tan\theta = \frac{P}{B}. \]
The last row reminds that the remaining three ratios are just the reciprocals of the first three: \[\text{cosec}\theta = \dfrac{1}{\sin\theta}\] \[\sec\theta = \dfrac{1}{\cos\theta}\] \[\cot\theta = \dfrac{1}{\tan\theta}\]
Why trigonometric ratios do NOT depend on the size of the triangle
Because any two right-angled triangles with the same angle have the same shape, even if one is bigger or smaller. Same shape \(\Rightarrow\) their sides increase or decrease in the same proportion.
So, although the lengths of the sides change, the ratio of those sides remains the same. That fixed ratio is what we call sin, cos, tan, etc.
Example-1
Given \(\tan A = \frac{4}{3}\), find the othertrigonometric ratios of the angle A.
Solution:
$$\begin{aligned}\tan A&=\dfrac{4}{3}\\\\ \tan A&=\dfrac{P}{B}\\ BC&=4k\quad\scriptsize\text{(k is a positive number)}\\ AB&=3k\\\\ AC&=\sqrt{\left( 4k\right) ^{2}+\left( 3k\right) ^{2}}\\ &=\sqrt{16k^{2}+9k^{2}}\\ &=\sqrt{25x^{2}}\\ &=5k\\\\ \sin A&=\dfrac{BC}{AC}\\ &=\dfrac{4k}{5k}\\ &=\dfrac{4}{5}\\\\ \cos A&=\dfrac{AB}{AC}\\ &=\dfrac{3k}{5k}\\ &=\dfrac{3}{5}\\\\ \text{cosec} A&=\dfrac{1}{\sin A}\\ &=\dfrac{1}{4/5}\\ &=\dfrac{5}{4}\\\\ \sec A&=\dfrac{1}{\cos A}\\ &=\dfrac{1}{3/5}\\ &=\dfrac{5}{3}\\\\ \cot A&=\dfrac{1}{\tan A}\\ &=\dfrac{1}{4/3}\\ &=\dfrac{3}{4}\end{aligned}$$Example-2
If \(\angle B\) and \(\angle Q\) are acute angles such that sin B = sin Q, then prove that \(\angle B = \angle Q.\)
Solution:
Given that sin B = sin Q
$$\begin{aligned}\sin B=\dfrac{AC}{AB}\\ \sin Q=\dfrac{PR}{PQ}\\ \sin B=\sin Q\\ \dfrac{AC}{AB}=\dfrac{PR}{PQ}=k\\ AC=kAB\\ PR=kPQ\\ BC=\sqrt{AB^{2}-AC^{2}}\\ QR=\sqrt{PQ^{2}-PR^{2}}\\ \dfrac{BC}{QR}=\dfrac{\sqrt{AB^{2}-AC^{2}}}{\sqrt{PQ^{2}-PR^{2}}}\\ \dfrac{BC}{QR}=\dfrac{\sqrt{\left( AB\right) ^{2}-\left( kAB\right) ^{2}}}{\sqrt{\left( PQ\right) ^{2}-\left( kPQ\right) ^{2}}}\\ \dfrac{BC}{QR}=\dfrac{AB\sqrt{1-k^{2}}}{PQ\sqrt{1-k^{2}}}\\ \dfrac{BC}{QR}=\dfrac{AB}{PQ}=\dfrac{AC}{PR}\\ \therefore \triangle ABC\sim \Delta PQR\left( SSS\right) \\ \Rightarrow \angle B=\angle Q\end{aligned}$$ Hence Proved.Example-3
Consider \(\triangle ACB\), right-angled at \(C\), in
which \(AB\) = 29 units, \(BC\) = 21 units and \(\angle ABC = \theta\)
(see Fig. 8.10). Determine the values of
(i) \(cos^2\ \theta + sin^2\ \theta\),
(ii) \(cos^2\ \theta – sin^2\ \theta\)
SOlutions:
AB = 29
BC = 21
Example-4
In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Solutions:
$$\begin{aligned}\tan A&=1\\\\ \tan A&=\dfrac{BC}{AB}\\ \dfrac{BC}{AB}&=1\\ BC&=AB\\ \text{Let}\\ BC=AB&=k\quad\scriptsize\text{(k is positive number)}\\\\ AC&=\sqrt{BC^{2}+AB^{2}}\\ &=\sqrt{k^{2}+k^{2}}\\ &=k\sqrt{2}\\\\ \sin\ A&=\dfrac{BC}{AC}\\ &=\dfrac{k}{k\sqrt{2}}\\ &=\dfrac{1}{\sqrt{2}}\\\\ \cos A&=\dfrac{AB}{AC}\\ &=\dfrac{k}{k\sqrt{2}}\\ &=\dfrac{1}{\sqrt{2}}\\\\ 2\times \sin A\cdot \cos A&=1\\\\ LHS\\\\ 2\times \dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}&=1\\\\ LHS&=RHS\end{aligned}$$ Hence ProvedExample-5
In \(\triangle OPQ\), right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q.
Solutions:
OP=7
OQ-PQ = 1
To find sin Q and cos Q
Trigonometric Ratios of Some Specific Angles
| Angle (°) | sin θ | cos θ | tan θ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | \(\frac{1}{2}\) | \(\frac{\sqrt3}{2}\) | \(\frac{1}{\sqrt3}\) |
| 45° | \(\frac{1}{\sqrt2}\) | \(\frac{1}{\sqrt2}\) | 1 |
| 60° | \(\frac{\sqrt3}{2}\) | \(\frac{1}{2}\) | \(\sqrt3\) |
| 90° | 1 | 0 | Not defined |
Tricks to remember Trigonometric Ratios of Some Specific Angles
Angles: \(0^\circ,\ 30^\circ,\ 45^\circ,\ 60^\circ,\ 90^\circ\)
Step 1 (for \(\sin\theta\)): Think of the numbers 0 1 2 3 4 in order.
Divide each by 4 and put a square root: \[ \sin\theta:\quad \sqrt{\tfrac{0}{4}},\ \sqrt{\tfrac{1}{4}},\ \sqrt{\tfrac{2}{4}},\ \sqrt{\tfrac{3}{4}},\ \sqrt{\tfrac{4}{4}} \]
These match \(\sin 0^\circ, \sin 30^\circ, \sin 45^\circ, \sin 60^\circ, \sin 90^\circ\) in this same order.
For \(\cos\theta\), use the same five values \(\sqrt{\tfrac{0}{4}}, \sqrt{\tfrac{1}{4}}, \sqrt{\tfrac{2}{4}}, \sqrt{\tfrac{3}{4}}, \sqrt{\tfrac{4}{4}}\) but read them from right to left for the angles \(0^\circ\) to \(90^\circ\).
So the \(\sin\) row goes from 0 to 4 under the root, and the \(\cos\) row goes from 4 back to 0 under the root. One small pattern helps you remember the full table for these special angles.
Example-6
In \(triangle ABC\), right-angled at B, AB = 5 cm and \(angle ACB = 30°\) (see Fig. 8.19). Determine the lengths of the sides BC and AC.
Solutions:
AB = 5cm, \(angle ACB = 30°\), to find AC and BC we can use any of the 3 ratio.
$$\begin{aligned}\sin 30^{\circ }&=\dfrac{AB}{AC}\\ \dfrac{1}{2}&=\dfrac{5}{AC}\\ &=5\times 2\\ &=10\end{aligned}$$Side AC and AB are known we can either use pythagoras theorem to calculate the third side or trigonometrical ratio
$$\begin{aligned}AC^{2}&=AB^{2}+BC^{2}\\ &=AC^{2}-AB^{2}\\ &=10^{2}-5^{2}\\ &=100-25\\ &=75\\ BC&=\sqrt{75}\\ &=5\times \sqrt{3}\end{aligned}$$Same Result can be acheived by using trigonometrical ratios
$$\begin{aligned}\cos 30^{\circ }&=\dfrac{BC}{AC}\\ \dfrac{\sqrt{3}}{2}&=\dfrac{BC}{10}\\ BC&=\dfrac{10\times \sqrt{3}}{2}\\ &=5\times \sqrt{3}\end{aligned}$$Example-7
In \(\triangle PQR\), right-angled at Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. Determine \(\angle QPR\) and \(\angle PRQ.\)
Solutions:
PQ = 3cm, PR = 6 cm
$$\begin{aligned}\sin R&=\dfrac{PQ}{PR}\\ &=\dfrac{3}{6}\\ &=\dfrac{1}{2}\\ \angle R&=30^{\circ }\end{aligned}$$By Angle Sum property
$$\begin{aligned}\angle P+\angle Q+\angle R&=180^{\circ}\\ \angle P+90^{\circ}+30^{\circ}&=180^{\circ }\\ \angle P&=180^{\circ }-120^{\circ }\\ \angle P&=60^{\circ}\end{aligned}$$Trigonometric Identities
- \(cos^2\ A + sin^2\ A =1\)
- \( 1 + tan^2\ A =sec^2\ A\)
- \(cot^2\ A + 1 =cosec^2\ A\)
Example-8
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
$$\begin{aligned} \cos ^{2}A+\sin ^{2}A&=1\\ \cos ^{2}A&=1-\sin ^{2}A\\ \cos A&=\pm \sqrt{1-\sin ^{2}A}\\ &=\sqrt{1-\sin ^{2}A}\\ &\scriptsize\left[ \because \left( -1 \lt \sin A \lt 1\right) \right] \end{aligned}$$ $$\begin{aligned}\sec A&=\dfrac{1}{\cos A}\\ \\ &=\dfrac{1}{\sqrt{1-\sin ^{2}A}}\\\\ \tan A&=\dfrac{\sin A}{\cos A}\\\\ &=\dfrac{\sin A}{\sqrt{1-\sin ^{2}A}}\end{aligned}$$Example-9
Prove that sec A (1 – sin A)(sec A + tan A) = 1
Solution:
$$\scriptsize\begin{aligned}LHS\\\\ &\sec A\left( 1-\sin A\right) \times \left( \sec A+\tan A\right) \\\\ &=\dfrac{1}{\cos A}\times \left( 1-\sin A\right) \times \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right) \\\\ &=\left( \dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}\right) \times \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right) \\\\ &=\left( \dfrac{1-\sin A}{\cos A}\right) \times \left( \dfrac{1+\sin A}{\cos A}\right) \\\\ &=\dfrac{1}{\cos ^{2}A}\times \left[ \left( 1-\sin A\right) \times \left( 1+\sin A\right) \right] \\\\ &=\dfrac{1}{\cos ^{2}A}\times \left[ 1-\sin ^{2}A\right] \\\\ &+\dfrac{\cos ^{2}A}{\cos ^{2}A}\\\\ &=1\\\\ LHS&=RHS\end{aligned}$$ Hence Proved.Example-10
Prove that \(\dfrac{\cot\ A–\cos\ A}{\cot\ A+\cos\ A}=\dfrac{\text{cosec }A-1}{\text{cosec }A+1}\)
Solution:
$$\begin{aligned}LHS\\\\ &\dfrac{\cot A-\cos A}{\cot A+\cos A}\\\\ &=\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}+\cos A}\\\\ &=\dfrac{\dfrac{\cos A-\cos A\times \sin A}{\sin A}}{\dfrac{\cos A+\cos A\times \sin A}{\sin A}}\\\\ &=\dfrac{\cos A-\cos A\cdot \sin A}{\cos A+\cos A\cdot \sin A}\\\\ &=\dfrac{\cos A\times \left( 1-\sin A\right) }{\cos A\times \left( 1+\sin A\right) }\\\\ &=\dfrac{1-\sin A}{1+\sin A}\end{aligned}$$Dividing numerator and Denominator by sin A
$$\begin{aligned}&=\dfrac{\dfrac{1-\sin A}{\sin A}}{\dfrac{1+\sin A}{\sin A}}\\\\ &=\dfrac{\dfrac{1}{\sin A}-1}{\dfrac{1}{\sin A}+1}\\\\ &=\dfrac{\text{cosec }A -1}{\text{cosec }A +1}\\\\ LHS&=RHS\end{aligned}$$ Hence ProvedExample-11
Prove that \(\dfrac{\sin\ \theta-\cos\ \theta +1}{\sin\ \theta +\cos\ \theta -1}=\dfrac{1}{\sec\ \theta - \tan\ \theta}\)
Solution:
$$\scriptsize\begin{aligned}LHS\\\\ &\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}\\\\ &=\dfrac{\cos \theta \times \left( \dfrac{\sin \theta }{\cos \theta }-\dfrac{\cos \theta }{\cos \theta }+\dfrac{1}{\cos \theta }\right) }{\cos \theta \times \left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\cos \theta }-\dfrac{1}{\cos \theta }\right) }\\\\ &=\dfrac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }\\\\ &=\dfrac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}\end{aligned}$$Multiply and divide by \((\tan \theta -\sec \theta )\)
$$\scriptsize\begin{aligned}&=\dfrac{\left( \tan \theta +\sec \theta -1\right) \times \left( \tan \theta -\sec \theta \right) }{\left( \tan \theta -\sec \theta +1\right) \times \left( \tan \theta -\sec \theta \right) }\\\\ &=\dfrac{\left( \tan ^{2}\theta -\sec ^{2}\theta \right) -1\times \left( \tan \theta -\sec \theta \right) }{\left( \tan \theta -\sec \theta +1\right) \times \left( \tan \theta -\sec \theta \right) }\\\\ &=\dfrac{\left( \tan ^{2}\theta -\left( 1+\tan ^{2}\theta \right) +\sec \theta -\tan \theta \right) }{\left( \tan \theta -\sec \theta +1\right) \times \left( \tan \theta -\sec \theta \right) }\\\\ &=\dfrac{\sec \theta -\tan \theta -1}{\left( \tan \theta -\sec \theta +1\right) \times \left( \tan \theta -\sec \theta \right) }\\\\ &=\dfrac{-\left( \tan \theta -\sec \theta +1\right) }{\left( \tan \theta -\sec \theta +1\right) \times \left( \tan \theta -\sec \theta \right) }\\\\ &=\dfrac{-1}{\tan \theta -\sec \theta }\\\\ &=\dfrac{1}{\sec \theta -\tan \theta }=RHS\\\\ LHS=RHS\end{aligned}$$ Hence ProvedImportant Points
- In a right triangle ABC, right-angled at B, \[ \begin{aligned} \sin\ A=\dfrac{\text{side opposite to angle A}}{\text{hypotenuse}}\\\\ \cos\ A=\dfrac{\text{side adjacent to angle A}}{\text{hypotenuse}}\\\\ \tan\ A=\dfrac{\text{side opposite toangle A}}{\text{side adjacent to angle A}} \end{aligned} \]
- \[ \begin{aligned} \text{cosec}\ A=\dfrac{1}{\sin\ A}\\\\ \sec\ A=\dfrac{1}{\cos\ A}\\\\ \tan\ A=\dfrac{1}{\cot\ A}\\\\ \tan\ A=\dfrac{\sin\ A}{\cos\ A} \end{aligned} \]
- If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of the angle can be easily determined.
- The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° A < 90°) or cosec A (0° < A 90º) is always greater than or equal to 1.
- \[\begin{aligned}\sin^2\ A + \cos^2\ A &= 1\\ \sec^2\ A – \tan^2\ A &= 1 \text{ for } 0° A < 90°\\ \text{cosec}^2\ A &=1 + cot^2\ A \text{ for } 0° < A 90º\end{aligned}\]
