💻✏Trigonometry Basics
1) Circular System Radian Measure
Statement: In the circular system of angle measurement, the unit is the radian. One radian is the angle subtended at the center of a circle by an arc whose length is equal to the radius.
\[
\theta = \frac{l}{r} \quad \text{(in radians)}
\]
where \( l \) = arc length, \( r \) = radius.
- \(1\ \text{right angle} = \frac{\pi}{2} \ \text{rad}\)
- \(1\ \text{straight angle} = \pi \ \text{rad}\)
- \(1\ \text{complete angle} = 2\pi \ \text{rad}\)
- Relation: \( 1^\circ = \frac{\pi}{180} \ \text{rad} \)
Careful: Remember that \( \pi \ \text{rad} = 180^\circ \), so \(\frac{\pi}{2} \ \text{rad} = 90^\circ\).
Convert Degrees to Radians
Example: Convert \(75^\circ\) to radians.
Solution:
\[ 75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \ \text{rad} \]Angle Conversion – Worked Variations Deg ⇄ Rad
Variation 1 (deg → rad): Convert \(135^\circ\) to radians.
\[
1^\circ=\frac{\pi}{180}\ \text{rad}
\]
\[
135^\circ = 135 \times \frac{\pi}{180}
= \frac{135}{180}\pi
= \frac{3\pi}{4}\ \text{rad}
\]
Convert Degrees + Minutes to Radians
Variation 2 (deg+minutes → rad): Convert \(-52^\circ 15'\) to radians.
\[
-52^\circ 15' = -\!\Big(52 + \frac{15}{60}\Big)^\circ
= -\!\Big(52 + \frac14\Big)^\circ
= -\frac{209}{4}^\circ
\]
\[
\text{rad} = -\frac{209}{4}\times\frac{\pi}{180}
= -\frac{209\pi}{720}\ \text{rad}
\]
Variation 3 (deg+min+sec → rad): Convert \(23^\circ 40' 30''\) to radians.
\[
23^\circ 40' 30'' = \left(23+\frac{40}{60}+\frac{30}{3600}\right)^\circ
= \left(23+\frac{2}{3}+\frac{1}{120}\right)^\circ
= \frac{947}{40}^\circ
\]
\[
\text{rad} = \frac{947}{40}\times\frac{\pi}{180}
= \frac{947\pi}{7200}\ \text{rad}
\]
Variation 4 (rad → deg+min): Convert \(\dfrac{5\pi}{24}\) rad to degrees–minutes.
\[
\text{deg} = \frac{5\pi}{24}\times\frac{180}{\pi}
= \frac{900}{24}^\circ
= 37.5^\circ
= 37^\circ 30'
\]
Variation 5 (rad → deg+min+sec): Convert \(\dfrac{5\pi}{14}\) rad to degrees–minutes–seconds.
\[
\text{deg} = \frac{5\pi}{14}\times\frac{180}{\pi}
= \frac{900}{14}^\circ
= \frac{450}{7}^\circ
= 64^\circ + \frac{2}{7}^\circ
\]
Convert the fractional degree to minutes:
\[
\frac{2}{7}^\circ \times 60 = \frac{120}{7}' = 17' + \frac{1}{7}'
\]
Convert the remaining fraction of a minute to seconds:
\[
\frac{1}{7}' \times 60 = \frac{60}{7}'' = 8\frac{4}{7}'' \approx 8.57''
\]
Final:
\[
\boxed{64^\circ\,17'\,8\frac{4}{7}'' \;\;(\approx 64^\circ\,17'\,8.57'')}
\]
Quick Answers:
(1) \(3\pi/4\) rad •
(2) \(-209\pi/720\) rad •
(3) \(947\pi/7200\) rad •
(4) \(37^\circ 30'\) •
(5) \(64^\circ 17' 8.57''\) (approx)
2) Trigonometric Ratios Right-Angled Triangle
Statement:
Trigonometric ratios (T-ratios) relate the angles of a right-angled triangle to the ratios of its sides.
Consider a right triangle \( \triangle ABO \) with:
- \(AB\) = perpendicular
- \(OB\) = base
- \(OA\) = hypotenuse
\[
\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{OA}
\]
\[
\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{OB}{OA}
\]
\[
\tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{OB}
\]
\[
\cot\theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{OB}{AB}
\]
\[
\sec\theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{OA}{OB}
\]
\[
\csc\theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{OA}{AB}
\]
Example: In a right triangle, \(AB = 3\), \(OB = 4\), find \(\sin\theta\) and \(\cos\theta\).
Solution:
Hypotenuse \(OA = \sqrt{3^2 + 4^2} = 5\). \[ \sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5} \]Question 1:
If \( \sin \theta = \frac{3}{5} \), where \( \theta \) lies in the first quadrant, find all the other trigonometric ratios. \[\textbf{Solution:}\] Let \( \triangle PQR \) be a right-angled triangle, right-angled at \( Q \).
\[ \sin \theta = \frac{3}{5} \quad \Rightarrow \quad \text{Hypotenuse (PR)} = 5, \quad \text{Opposite side (PQ)} = 3 \] On applying the Pythagorean theorem in \( \triangle PQR \), we get: \[ (\text{Hypotenuse})^2 = (\text{Opposite})^2 + (\text{Adjacent})^2 \] \[ 5^2 = 3^2 + (\text{Adjacent})^2 \quad \Rightarrow \quad 25 = 9 + (\text{Adjacent})^2 \] \[ \text{Adjacent} = \sqrt{25 - 9} = \sqrt{16} = 4 \] Now, we calculate all the trigonometric ratios: \[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5} \] \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \] \[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{4}{3} \] \[ \sec \theta = \frac{1}{\cos \theta} = \frac{5}{4} \] \[ \csc \theta = \frac{1}{\sin \theta} = \frac{5}{3} \]
Question 2:
If \( \cos \theta = \frac{5}{13} \), where \( \theta \) lies in the first quadrant, find all the other trigonometric ratios.
\[\textbf{Solution:}\] Let \( \triangle PQR \) be a right-angled triangle, right-angled at \( Q \). \[ \cos \theta = \frac{5}{13} \quad \Rightarrow \quad \text{Adjacent side (PQ)} = 5, \quad \text{Hypotenuse (PR)} = 13 \] On applying the Pythagorean theorem in \( \triangle PQR \), we get: \[ (\text{Hypotenuse})^2 = (\text{Adjacent})^2 + (\text{Opposite})^2 \] \[ 13^2 = 5^2 + (\text{Opposite})^2 \quad \Rightarrow \quad 169 = 25 + (\text{Opposite})^2 \] \[ \text{Opposite} = \sqrt{169 - 25} = \sqrt{144} = 12 \] Now, we calculate all the trigonometric ratios: \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13} \] \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{12}{5} \] \[ \cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{5}{12} \] \[ \sec \theta = \frac{1}{\cos \theta} = \frac{13}{5} \] \[ \csc \theta = \frac{1}{\sin \theta} = \frac{13}{12} \]
Question 3:
In \( \triangle XYZ \), \( m \angle X = \frac{3\pi}{4} \) and \( m \angle Y = 30^\circ \). Find \( m \angle Z \) in both systems.
\[ \textbf{Solution:} \] We are given that \[ m \angle X = \frac{3\pi}{4} \quad \text{(in radians)} \] Converting to degrees: \[ m \angle X = \left( \frac{3\pi}{4} \times \frac{180^\circ}{\pi} \right) = 135^\circ \] And we are given that: \[ m \angle Y = 60^\circ \] Now, in \( \triangle XYZ \), the sum of the angles is \( 180^\circ \). Thus: \[ m \angle X + m \angle Y + m \angle Z = 180^\circ \] Substitute the known values: \[ 135^\circ + 30^\circ + m \angle Z = 180^\circ \] \[ 165^\circ + m \angle Z = 180^\circ \] \[ m \angle Z = 180^\circ - 165^\circ \] \[ m \angle Z = 15^\circ \] To express in radians: \[ m \angle Z = 15^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{12} \] Thus, \[ m \angle Z = 15^\circ \quad \text{or} \quad m \angle Z = \frac{\pi}{12} \quad \text{(in radians)}. \]Question 4:
In \( \triangle ABC \), \( m \angle A = \frac{\pi}{6} \) and \( m \angle B = 45^\circ \). Find \( m \angle C \) in both systems.
\[ \textbf{Solution:} \]We are given that \[ m \angle A = \frac{\pi}{6} \quad \text{(in radians)} \] Converting to degrees: \[ m \angle A = \left( \frac{\pi}{6} \times \frac{180^\circ}{\pi} \right) = 30^\circ \] And we are given that: \[ m \angle B = 45^\circ \] Now, in \( \triangle ABC \), the sum of the angles is \( 180^\circ \). Thus: \[ m \angle A + m \angle B + m \angle C = 180^\circ \] Substitute the known values: \[ 30^\circ + 45^\circ + m \angle C = 180^\circ \] \[ 75^\circ + m \angle C = 180^\circ \] \[ m \angle C = 180^\circ - 75^\circ \] \[ m \angle C = 105^\circ \] To express in radians: \[ m \angle C = 105^\circ \times \frac{\pi}{180^\circ} = \frac{7\pi}{12} \] Thus, \[ m \angle C = 105^\circ \quad \text{or} \quad m \angle C = \frac{7\pi}{12} \quad \text{(in radians)}. \]
Question 5 :
Problem A (Arc length from degrees & minutes)
Question. In a circle of radius \(7\text{ cm}\), find the length of the arc that subtends an angle of \(48^\circ 30'\) at the center. (Use \( \pi=\dfrac{22}{7}\).)
Solution.
\[ \begin{aligned} r &= 7\ \text{cm}, \\ 48^\circ 30' &= 48^\circ + \frac{30}{60}^\circ = 48.5^\circ = \frac{97}{2}^\circ,\\[4pt] \theta &= \frac{97}{2}\cdot\frac{\pi}{180} = \frac{97\pi}{360}\ \text{radians},\\[6pt] \ell &= r\theta = 7\cdot\frac{97\pi}{360} = \frac{679\pi}{360}\ \text{cm},\\[6pt] \text{Using }\pi &\approx \frac{22}{7}:\quad \ell \approx \frac{679\cdot 22}{360\cdot 7} = \frac{1067}{180} \approx 5.93\ \text{cm}. \end{aligned} \]Question 6 :
Problem B (Minute hand travel)
Question. The minute hand of a clock is \(12\text{ cm}\) long. How far does its tip move in \(25\) minutes? (Use \( \pi=\dfrac{22}{7}\).)
Solution.
\[ \begin{aligned} \text{In }60\text{ min: } & 2\pi\ \text{radians}.\\ \text{In }25\text{ min: } & \theta = 2\pi\cdot\frac{25}{60} = \frac{5\pi}{6}.\\[6pt] \ell &= r\theta = 12\cdot\frac{5\pi}{6} = 10\pi\ \text{cm}.\\[6pt] \text{Using }\pi &\approx \frac{22}{7}:\quad \ell \approx \frac{220}{7} \approx 31.43\ \text{cm}. \end{aligned} \]Question 7 :
Problem A (Fourth angle of a quadrilateral)
Question. If three angles of a quadrilateral are \(75^\circ\), \(80^{\mathrm g}\) (grads), and \(\dfrac{2\pi}{5}\) radians, find the fourth angle.
Solution.
\[ \begin{aligned} \text{Sum of interior angles of a quadrilateral} &= 360^\circ.\\[4pt] 80^{\mathrm g} &= 80\times \frac{90}{100} = 72^\circ.\\[4pt] \frac{2\pi}{5}\ \text{rad} &= \frac{2\pi}{5}\cdot \frac{180^\circ}{\pi} = 72^\circ.\\[4pt] \text{Sum of known angles} &= 75^\circ + 72^\circ + 72^\circ = 219^\circ.\\[4pt] \text{Fourth angle} &= 360^\circ - 219^\circ = \boxed{141^\circ}. \end{aligned} \]Question 7 :
Problem B (Arc length from chord)
Question. In a circle of diameter \(20\ \text{cm}\), a chord has length \(10\sqrt{3}\ \text{cm}\). Find the length of the minor arc subtended by the chord.
Solution.
\[ \begin{aligned} R &= \frac{20}{2} = 10\ \text{cm}.\\[4pt] \text{Chord formula:}\quad c &= 2R\sin\!\left(\frac{\theta}{2}\right).\\[4pt] \sin\!\left(\frac{\theta}{2}\right) &= \frac{c}{2R} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2} \ \Rightarrow\ \frac{\theta}{2} = 60^\circ = \frac{\pi}{3}.\\[4pt] \theta &= 120^\circ = \frac{2\pi}{3}\ \text{rad}.\\[6pt] \text{Arc length:}\quad s &= R\theta = 10\cdot \frac{2\pi}{3} = \boxed{\frac{20\pi}{3}\ \text{cm}}. \end{aligned} \]3) Signs of Trigonometric Ratios Signs in Various Quadrants
Trigonometric Identities Formulas & Signs
Statement: Trigonometric identities are equations true for all angles where both sides are defined. Use identities together with the quadrant sign rules to evaluate or simplify expressions quickly.
Pythagorean identities
\[
\sin^2\theta + \cos^2\theta = 1,\qquad
1+\tan^2\theta = \sec^2\theta,\qquad
1+\cot^2\theta = \csc^2\theta.
\]
Reciprocal & quotient identities
\[
\sin\theta=\frac{1}{\csc\theta},\quad
\cos\theta=\frac{1}{\sec\theta},\quad
\tan\theta=\frac{\sin\theta}{\cos\theta},\quad
\cot\theta=\frac{\cos\theta}{\sin\theta}.
\]
Cofunction (complements)
\[
\sin(90^\circ-\theta)=\cos\theta,\quad
\cos(90^\circ-\theta)=\sin\theta,\quad
\tan(90^\circ-\theta)=\cot\theta,
\]
\[
\csc(90^\circ-\theta)=\sec\theta,\quad
\sec(90^\circ-\theta)=\csc\theta,\quad
\cot(90^\circ-\theta)=\tan\theta.
\]
Even–odd & periodicity
\[
\sin(-\theta)=-\sin\theta,\quad
\cos(-\theta)=\cos\theta,\quad
\tan(-\theta)=-\tan\theta,\quad
\cot(-\theta)=-\cot\theta;
\]
\[
\sin(\theta+2\pi)=\sin\theta,\quad
\cos(\theta+2\pi)=\cos\theta,\quad
\tan(\theta+\pi)=\tan\theta,\quad
\cot(\theta+\pi)=\cot\theta.
\]
Signs by Quadrant (ASTC): Use the reference angle \( \alpha \) (the acute angle to the nearest axis). Decide the sign from the quadrant, then apply identities to the positive value at \( \alpha \).
- I (0°–90°): all \( \sin,\cos,\tan,\csc,\sec,\cot \) are \(+\).
- II (90°–180°): \( \sin,\csc \) are \(+\); \( \cos,\sec,\tan,\cot \) are \( - \).
- III (180°–270°): \( \tan,\cot \) are \(+\); \( \sin,\csc,\cos,\sec \) are \( - \).
- IV (270°–360°): \( \cos,\sec \) are \(+\); \( \sin,\csc,\tan,\cot \) are \( - \).
- Reference-angle rule: For any \( \theta \), compute \( \alpha = \big|\theta\big|\) mod \(90^\circ\), find the quadrant of \( \theta \), use the sign from ASTC, and evaluate the function at \( \alpha \).
- Identity + sign = speed: Reduce using identities first, then assign the sign via quadrant.
Example 1: Evaluate \( \sin 150^\circ \).
Solution:
\(150^\circ\) is in Quadrant II ⇒ sine is \(+\). Reference angle \( \alpha=30^\circ\). \[ \sin 150^\circ = +\sin 30^\circ = \frac{1}{2}. \]
Example 2: Evaluate \( \cos 210^\circ \).
Solution:
\(210^\circ\) is in Quadrant III ⇒ cosine is \( - \). Reference angle \( \alpha=30^\circ\). \[ \cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}. \]
Example 3: Simplify \( \sec^2\theta - \tan^2\theta \).
Solution:
From \(1+\tan^2\theta=\sec^2\theta\), \[ \sec^2\theta - \tan^2\theta = 1. \] (True in any quadrant where both sides are defined.)
Careful: Identities give magnitudes/relations; the sign still depends on the quadrant.
Also watch for undefined points (e.g., \( \tan\theta \) undefined when \( \cos\theta=0 \)).
4) T-ratios of Allied Angles Allied angles is an important concerpt
\[
\textbf{T\text{-}ratios of Allied Angles:} \quad
\text{Two angles are allied if their sum or difference is a multiple of } 90^\circ.
\]
- Case (i): When angle \( \theta \) is negative
- \(\sin(-\theta) = -\sin\theta\)
- \(\cos(-\theta) = \cos\theta\)
- \(\tan(-\theta) = -\tan\theta\)
- Case (ii): When angle \( \theta \) is less than \(90^\circ\) (i.e., lies in I quadrant)
- \(\sin(90^\circ - \theta) = \cos\theta\)
- \(\cos(90^\circ - \theta) = \sin\theta\)
- \(\tan(90^\circ - \theta) = \cot\theta\)
- Case (iii): When angle \( \theta \) lies between \(90^\circ\) and \(180^\circ\) (i.e., in II quadrant)
- \(\sin(90^\circ + \theta) = \cos\theta\)
- \(\cos(90^\circ + \theta) = -\sin\theta\)
- \(\tan(90^\circ + \theta) = -\cot\theta\)
- \(\sin(180^\circ - \theta) = \sin\theta\)
- \(\cos(180^\circ - \theta) = -\cos\theta\)
- \(\tan(180^\circ - \theta) = -\tan\theta\)
- Case (iv): When angle \( \theta \) lies between \(180^\circ\) and \(270^\circ\) (i.e., in III quadrant)
- \(\sin(180^\circ + \theta) = -\sin\theta\)
- \(\cos(180^\circ + \theta) = -\cos\theta\)
- \(\tan(180^\circ + \theta) = \tan\theta\)
- \(\sin(270^\circ - \theta) = -\cos\theta\)
- \(\cos(270^\circ - \theta) = -\sin\theta\)
- \(\tan(270^\circ - \theta) = \cot\theta\)
\[
\textbf{T-ratios of Allied Angles:}
\quad {\text{Two angles are allied if their sum/difference is a multiple of }90^\circ.}
\]
- Case (i): When angle \( \theta \) is negative
- \(\sin(-\theta)= -\sin\theta\)
- \(\cos(-\theta)= \cos\theta\)
- \(\tan(-\theta)= -\tan\theta\)
- Case (ii): When \( \theta < 90^\circ \) (I quadrant)
- \(\sin(90^\circ-\theta)=\cos\theta\)
- \(\cos(90^\circ-\theta)=\sin\theta\)
- \(\tan(90^\circ-\theta)=\cot\theta\)
- Case (iii): When \(90^\circ<\theta<180^\circ\) (II quadrant)
- \(\sin(90^\circ+\theta)=\cos\theta\)
- \(\cos(90^\circ+\theta)=-\sin\theta\)
- \(\tan(90^\circ+\theta)=-\cot\theta\)
- \(\sin(180^\circ-\theta)=\sin\theta\)
- \(\cos(180^\circ-\theta)=-\cos\theta\)
- \(\tan(180^\circ-\theta)=-\tan\theta\)
- Case (iv): When \(180^\circ<\theta<270^\circ\) (III quadrant)
- \(\sin(180^\circ+\theta)=-\sin\theta\)
- \(\cos(180^\circ+\theta)=-\cos\theta\)
- \(\tan(180^\circ+\theta)=\tan\theta\)
- \(\sin(270^\circ-\theta)=-\cos\theta\)
- \(\cos(270^\circ-\theta)=-\sin\theta\)
- \(\tan(270^\circ-\theta)=\cot\theta\)
- Case (v): When \(270^\circ<\theta<360^\circ\) (IV quadrant)
- \(\sin(270^\circ+\theta)=-\cos\theta\)
- \(\cos(270^\circ+\theta)=\sin\theta\)
- \(\tan(270^\circ+\theta)=-\cot\theta\)
- \(\sin(360^\circ-\theta)=-\sin\theta\)
- \(\cos(360^\circ-\theta)=\cos\theta\)
- \(\tan(360^\circ-\theta)=-\tan\theta\)
\[
\textbf{Values of T-ratios for some standard angles}
\]
| Angle \( \theta \) | \(0^\circ\) (\(0\)) |
\(30^\circ\) (\(=\frac{\pi}{6}\)) |
\(45^\circ\) (\(=\frac{\pi}{4}\)) |
\(60^\circ\) (\(=\frac{\pi}{3}\)) |
\(90^\circ\) (\(=\frac{\pi}{2}\)) |
\(120^\circ\) (\(=\frac{2\pi}{3}\)) |
\(135^\circ\) (\(=\frac{3\pi}{4}\)) |
\(150^\circ\) (\(=\frac{5\pi}{6}\)) |
\(180^\circ\) (\(=\pi\)) |
|---|---|---|---|---|---|---|---|---|---|
| \(\sin\theta\) | \(0\) | \(\tfrac{1}{2}\) | \(\tfrac{1}{\sqrt{2}}\) | \(\tfrac{\sqrt{3}}{2}\) | \(1\) | \(\tfrac{\sqrt{3}}{2}\) | \(\tfrac{1}{\sqrt{2}}\) | \(\tfrac{1}{2}\) | \(0\) |
| \(\cos\theta\) | \(1\) | \(\tfrac{\sqrt{3}}{2}\) | \(\tfrac{1}{\sqrt{2}}\) | \(\tfrac{1}{2}\) | \(0\) | \(-\tfrac{1}{2}\) | \(-\tfrac{1}{\sqrt{2}}\) | \(-\tfrac{\sqrt{3}}{2}\) | \(-1\) |
| \(\tan\theta\) | \(0\) | \(\tfrac{1}{\sqrt{3}}\) | \(1\) | \(\sqrt{3}\) | \(\infty\) | \(-\sqrt{3}\) | \(-1\) | \(-\tfrac{1}{\sqrt{3}}\) | \(0\) |
5) Solved Examples Study and practice the steps
\[
\textbf{Worked Examples with Allied-Angle Identities}
\quad\text{(rewrite angles to } 0^\circ,90^\circ,180^\circ,270^\circ,360^\circ\text{ and use exact values).}
\]
\[
\textbf{Example 1: } \ \sin(-30^\circ)
\]
\[
\begin{aligned}
\sin(-30^\circ) &= -\sin 30^\circ
&& \text{[∵ } \sin(-\theta)=-\sin\theta \text{]}\\
&= -\tfrac12.
\end{aligned}
\]
\[
\textbf{Example 2: } \ \tan 300^\circ
\]
\[
\begin{aligned}
\tan 300^\circ &= \tan(270^\circ+30^\circ) && \text{(rewrite)}\\
&= -\cot 30^\circ && \text{[∵ } \tan(270^\circ+\theta)=-\cot\theta \text{]}\\
&= -\sqrt{3}.
\end{aligned}
\]
\[
\textbf{Example 3: } \ \cos 210^\circ
\]
\[
\begin{aligned}
\cos 210^\circ &= \cos(180^\circ+30^\circ)\\
&= -\cos 30^\circ && \text{[∵ } \cos(180^\circ+\theta)=-\cos\theta \text{]}\\
&= -\tfrac{\sqrt{3}}{2}.
\end{aligned}
\]
\[
\textbf{Example 4: } \ \sec 150^\circ
\]
\[
\begin{aligned}
\sec 150^\circ &= \sec(180^\circ-30^\circ)\\
&= -\sec 30^\circ && \text{[∵ } \sec(180^\circ-\theta)=-\sec\theta \text{]}\\
&= -\frac{1}{\cos 30^\circ}
= -\frac{1}{\tfrac{\sqrt{3}}{2}}
= -\frac{2}{\sqrt{3}}
= -\frac{2\sqrt{3}}{3}\,.
\end{aligned}
\]
\[
\textbf{Example 5: } \ \cot 135^\circ
\]
\[
\begin{aligned}
\cot 135^\circ &= \cot(90^\circ+45^\circ)\\
&= -\tan 45^\circ && \text{[∵ } \cot(90^\circ+\theta)=-\tan\theta \text{]}\\
&= -1.
\end{aligned}
\]
6) Some important formulae of trigonometryPractice these
- \[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]
- \[ \sin(A-B)=\sin A\cos B-\cos A\sin B \]
- \[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]
- \[ \cos(A-B)=\cos A\cos B+\sin A\sin B \]
- \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\,\tan B} \]
- \[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B} \]
- \[ \sin 2A=2\sin A\cos A=\frac{2\tan A}{1+\tan^2 A} \]
- \[ \cos 2A=\cos^2\!A-\sin^2\!A =2\cos^2\!A-1 =1-2\sin^2\!A =\frac{1-\tan^2\!A}{1+\tan^2\!A} \]
- \[ \tan 2A=\frac{2\tan A}{1-\tan^2 A} \]
- \[ \sin 3A=3\sin A-4\sin^3\!A \]
- \[ \cos 3A=4\cos^3\!A-3\cos A \]
More important formulae of trigonometry
- \[ \tan 3A \;=\; \frac{3\tan A - \tan^{3}\!A}{\,1 - 3\tan^{2}\!A\,} \]
- \[ \sin(A+B) + \sin(A-B) \;=\; 2\sin A\,\cos B \]
- \[ \sin(A+B) - \sin(A-B) \;=\; 2\cos A\,\sin B \]
- \[ \cos(A+B) + \cos(A-B) \;=\; 2\cos A\,\cos B \]
- \[ \cos(A+B) - \cos(A-B) \;=\; -\,2\sin A\,\sin B \]
- \[ \sin C + \sin D \;=\; 2\sin\!\left(\frac{C+D}{2}\right) \cos\!\left(\frac{C-D}{2}\right) \]
- \[ \sin C - \sin D \;=\; 2\cos\!\left(\frac{C+D}{2}\right) \sin\!\left(\frac{C-D}{2}\right) \]
- \[ \cos C + \cos D \;=\; 2\cos\!\left(\frac{C+D}{2}\right) \cos\!\left(\frac{C-D}{2}\right) \]
- \[ \cos C - \cos D \;=\; -\,2\sin\!\left(\frac{C+D}{2}\right) \sin\!\left(\frac{C-D}{2}\right) \]
7) Some Solved problemsWork these out
\[
\textbf{Worked Examples (using } \sin(A\pm B),\ \cos(A\pm B),\ \tan(A\pm B)\text{).}
\]
\[
\textbf{Example 1: Find } (i)\ \sin 75^\circ \quad (ii)\ \tan 15^\circ
\]
\[
\underline{(i)\ \sin 75^\circ}
\]
\[
\begin{aligned}
\sin 75^\circ &= \sin(45^\circ+30^\circ) \\
&= \sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
&& \text{[by } \sin(A{+}B)\text{]}\\
&= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}
\;+\; \frac{1}{\sqrt{2}}\cdot\frac{1}{2} \\
&= \frac{\sqrt{3}+1}{2\sqrt{2}}
= \frac{\sqrt{6}+\sqrt{2}}{4}.
\end{aligned}
\]
\[
\underline{(ii)\ \tan 15^\circ}
\]
\[
\begin{aligned}
\tan 15^\circ &= \tan(45^\circ-30^\circ) \\
&= \frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\,\tan30^\circ}
&& \text{[by } \tan(A{-}B)\text{]}\\
&= \frac{1-\tfrac{1}{\sqrt{3}}}{1+\tfrac{1}{\sqrt{3}}}
= \frac{\sqrt{3}-1}{\sqrt{3}+1} \\
&= \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}
= \frac{3-2\sqrt{3}+1}{3-1}
= \frac{4-2\sqrt{3}}{2}
= 2-\sqrt{3}.
\end{aligned}
\]
\[
\textbf{Example 2: Find } (i)\ \cos 75^\circ \quad (ii)\ \cot 15^\circ
\]
\[
\underline{(i)\ \cos 75^\circ}
\]
\[
\begin{aligned}
\cos 75^\circ &= \cos(45^\circ+30^\circ) \\
&= \cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ
&& \text{[by } \cos(A{+}B)\text{]}\\
&= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}
\;-\; \frac{1}{\sqrt{2}}\cdot\frac{1}{2} \\
&= \frac{\sqrt{3}-1}{2\sqrt{2}}
= \frac{\sqrt{6}-\sqrt{2}}{4}.
\end{aligned}
\]
\[
\underline{(ii)\ \cot 15^\circ}
\]
\[
\begin{aligned}
\cot 15^\circ &= \frac{1}{\tan 15^\circ}
= \frac{1}{\,2-\sqrt{3}\,}
= \frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} \\
&= 2+\sqrt{3}.
\end{aligned}
\]
8) Trigonometric-Identities- Solved problemsWork these out
\[ \textbf{ Show that}\]
\[
2(\sin^6x + \cos^6x) - 3(\sin^4x + \cos^4x) + 1 = 0
\]
\[\textbf{Solution:}\]
Step 1: Rewrite in terms of squares.
\[
2(\sin^6x + \cos^6x) - 3(\sin^4x + \cos^4x) + 1
= 2\big[(\sin^2x)^3 + (\cos^2x)^3\big] - 3\big[(\sin^2x)^2 + (\cos^2x)^2\big] + 1
\]
\[
\textbf{Step 2: Use the identity}
\]
\[
a^3 + b^3 = (a+b)^3 - 3ab(a+b).
\]
\[
a = sin^2x, b = cos^2x\]
\[
(\sin^2x)^3 + (\cos^2x)^3 = (1)^3 - 3\sin^2x\cos^2x = 1 - 3\sin^2x\cos^2x
\]
\[
\textbf{Step 3: Use the identity}
\]
\[
a^2 + b^2 = (a+b)^2 - 2ab
\]
\[
(\sin^2x)^2 + (\cos^2x)^2 = (1)^2 - 2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x
\]
Step 4: Substitute these results:
\[
2[1 - 3\sin^2x\cos^2x] - 3[1 - 2\sin^2x\cos^2x] + 1
\]
Step 5: Simplify:
\[
= (2 - 3 + 1) + (-6\sin^2x\cos^2x + 6\sin^2x\cos^2x) = 0
\]
Thus proved.
\[ \textbf{Show that}\]
\[
\text{(i) } \sin^8A - \cos^8A = (\sin^2A - \cos^2A)(1 - 2\sin^2A \cos^2A)
\]
\[\textbf{Solution (i):}\]
Step 1: Start with the left-hand side.
\[
\sin^8A - \cos^8A = (\sin^4A)^2 - (\cos^4A)^2
\]
Step 2: Use the difference of squares identity.
\[
(\sin^4A)^2 - (\cos^4A)^2 = (\sin^4A - \cos^4A)(\sin^4A + \cos^4A)
\]
Step 3: Again use the difference of squares on the first factor.
\[
\sin^4A - \cos^4A = (\sin^2A - \cos^2A)(\sin^2A + \cos^2A)
\]
Step 4: Expand the second factor.
\[
\sin^4A + \cos^4A = (\sin^2A + \cos^2A)^2 - 2\sin^2A \cos^2A
\]
Step 5: Put everything together.
\[
\sin^8A - \cos^8A = (\sin^2A - \cos^2A)(\sin^2A + \cos^2A)\big[(\sin^2A + \cos^2A)^2 - 2\sin^2A \cos^2A\big]
\]
Step 6: Since \[\sin^2A + \cos^2A = 1\],
\[
= (\sin^2A - \cos^2A)(1 - 2\sin^2A \cos^2A)
\]
Thus proved.
\[ \textbf{Example: Show that}\]
\[
\frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A}
\]
\[\textbf{Solution (ii):}\]
Step 1: Start with the given expression.
\[
\frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A}
\]
Step 2: Rearragne the equation.
\[
\frac{1}{\sec A - \tan A} - \frac{1}{\sec A + \tan A} = \frac{1}{\cos A} + \frac{1}{\cos A}
\]
Step 3: consider LHS.
\[
= \frac{1}{\sec A - \tan A} + \frac{1}{\sec A + \tan A}
\]
Step 4: consider lcm and simplify.
\[
= \frac{\sec A + \tan A + \sec A - \tan B}{(\sec A - \tan A)(\sec A + \tan A)}
\]
simplify
\[
= \frac{2\sec A}{\sec ^2A - \tan^2 A}
\]
\[
= 2\sec A
\]
\[
= \frac{2}{\cos A}
\]
Step 5: consider RHS.
\[
= \frac{1}{\cos A} + \frac{1}{\cos A}
\]
Step 6: Adding we get.
\[
= \frac{2}{\cos A}
\]
Step 5: Compare with the left-hand side.
Thus LHS = RHS, hence proved.
\[ \textbf{Example. If } \frac{\cos^4A}{\cos^2B} + \frac{\sin^4A}{\sin^2B} = 1,
\text{ then prove that} \]
\[
\text{(i) } \sin^4A + \sin^4B = 2\sin^2A \sin^2B
\]
\[
\text{(ii) } \frac{\cos^4B}{\cos^2A} + \frac{\sin^4B}{\sin^2A} = 1
\]
\[\textbf{Solution:}\]
Step 1: Start with the given relation.
\[
\frac{\cos^4A}{\cos^2B} + \frac{\sin^4A}{\sin^2B} = 1
\]
Since \[\cos^2A + \sin^2A = 1\], we can write:
\[
\frac{\cos^4A}{\cos^2B} + \frac{\sin^4A}{\sin^2B} = 1
\]
as
\[
\frac{\cos^4A}{\cos^2B} + \frac{\sin^4A}{\sin^2B} = \cos^2A + \sin^2A
\]
---
Step 2: Rearrange terms.
\[
\frac{\cos^4A}{\cos^2B} - \cos^2A = \sin^2A - \frac{\sin^4A}{\sin^2B}
\]
---
Step 3: Factorize each side.
\[
\frac{\cos^2A(\cos^2A - \cos^2B)}{\cos^2B}
= \frac{\sin^2A(\sin^2B - \sin^2A)}{\sin^2B}
\]
---
Step 4: Simplify.
\[
\frac{\cos^2A}{\cos^2B}(\cos^2A - \cos^2B)
= \frac{\sin^2A}{\sin^2B}(\cos^2A - \cos^2B)
\]
---
Step 5: Factor out \[(\cos^2A - \cos^2B)\]
\[
(\cos^2A - \cos^2B)\left(\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B}\right) = 0
\]
---
Step 6: Consider the two cases.
**Case 1:**
\[
\cos^2A - \cos^2B = 0 \quad \implies \quad \cos^2A = \cos^2B \tag{i}
\]
**Case 2:**
\[
\frac{\cos^2A}{\cos^2B} - \frac{\sin^2A}{\sin^2B} = 0
\quad \implies \quad \cos^2A \sin^2B = \sin^2A \cos^2B
\]
\[
\implies \cos^2A = \cos^2B \tag{ii}
\]
Thus in both cases,
\[
\cos^2A = \cos^2B \quad \implies \quad \sin^2A = \sin^2B \tag{iii}
\]
---
### Part (i)
Step 1: Take LHS.
\[
\text{LHS} = \sin^4A + \sin^4B
\]
Step 2: Expand using the identity \[(p+q)^2 = p^2 + 2pq + q^2.\]
\[
= (\sin^2A - \sin^2B)^2 + 2\sin^2A\sin^2B
\]
Step 3: Since \[\sin^2A = \sin^2B\] (from (iii)), the first term vanishes.
\[
= 2\sin^2A\sin^2B
\]
Step 4: Which equals RHS.
\[
\sin^4A + \sin^4B = 2\sin^2A \sin^2B
\]
Thus proved.
---
### Part (ii)
Step 1: Take LHS.
\[
\text{LHS} = \frac{\cos^4B}{\cos^2A} + \frac{\sin^4B}{\sin^2A}
\]
Step 2: Using $\cos^2A = \cos^2B$ and $\sin^2A = \sin^2B$,
\[
= \frac{\cos^4B}{\cos^2B} + \frac{\sin^4B}{\sin^2B}
\]
Step 3: Simplify.
\[
= \cos^2B + \sin^2B
\]
Step 4: Since \[\cos^2B + \sin^2B = 1\],
\[
\text{LHS} = 1 = \text{RHS}
\]
Thus proved.
\[ \textbf{Example: If } \tan^2\theta = 1 - e^2, \text{ prove that } \]
\[
\sec\theta + \tan^3\theta \csc\theta = (2 - e^2)^{\tfrac{3}{2}}
\]
\[\textbf{Solution:}\]
Step 1: Start with the Left Hand Side (LHS).
\[
\text{LHS} = \sec\theta + \tan^3\theta \csc\theta
\]
\[
= \sec\theta \left( 1 + \tan^3\theta \cdot \frac{\csc\theta}{\sec\theta} \right)
\]
Step 2:
\[
= \sec\theta \left( 1 + \tan^3\theta \cdot \cot\theta \right)
\]
Step 3: Simplify
\[
\tan^3\theta \cdot \cot\theta = \tan^2\theta
\]
\[
= \sec\theta (1 + \tan^2\theta)
\]
Step 4: Recall Identity
\[
1 + \tan^2\theta = \sec^2\theta
\]
\[
= \sec\theta \cdot \sec^2\theta = \sec^3\theta
\]
Step 5:
\[
\sec^3\theta = (\sec^2\theta)^{\tfrac{3}{2}}
\]
\[
= (1 + \tan^2\theta)^{\tfrac{3}{2}}
\]
Step 6: Substitute
\[
\tan^2\theta = 1 - e^2
\]
\[
= (1 + 1 - e^2)^{\tfrac{3}{2}} = (2 - e^2)^{\tfrac{3}{2}}
\]
Thus,
\[
\sec\theta + \tan^3\theta \csc\theta = (2 - e^2)^{\tfrac{3}{2}}
\]
Hence proved.
\[ \textbf{Example : Show that the equation } \sin\theta = x + \frac{1}{x} \text{ is impossible if $x$ is real.} \]
\[\textbf{Solution:}\]
Step 1: Suppose
\[
\sin\theta = x + \frac{1}{x}.
\]
Step 2: Square both sides.
\[
\sin^2\theta = \left(x + \frac{1}{x}\right)^2
\]
\[
= x^2 + \frac{1}{x^2} + 2\left(x \cdot \frac{1}{x}\right)
\]
\[
= x^2 + \frac{1}{x^2} + 2
\]
\[
\text{Step 3: Notice that } x^2 + \frac{1}{x^2} \geq 2 \; (\text{by AM–GM inequality})
\]
So,
\[
\sin^2\theta \geq 2 + 2 = 4
\]
\[
\text{Step 4: But } \sin^2\theta \leq 1 \text{ for all real } \theta
\]
\[
\text{This is a contradiction. Hence the given equation is impossible for real } x
\]