Vardaan Learning Institute
In elementary geometry, an angle is considered as the figure formed by two rays meeting at a common endpoint (vertex). In trigonometry, an angle is conceived as the amount of rotation of a revolving ray from an initial position to a terminal position about a fixed vertex.
1. Sexagesimal System (English System / Degree Measure):
2. Circular System (Radian Measure):
A radian is the measure of an angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle. It is denoted by $1^{\text{c}}$ or simply $1 \text{ rad}$.
Fundamental Relationship: The circumference of a circle of radius $r$ is $2\pi r$. Therefore, one complete revolution ($360^\circ$) subtends an angle of $\dfrac{2\pi r}{r} = 2\pi\text{ radians}$ at the centre.
$$\pi\text{ radians} = 180^\circ \iff 1\text{ rad} = \left(\frac{180}{\pi}\right)^\circ \approx 57^\circ 16' 22''$$ $$1^\circ = \left(\frac{\pi}{180}\right)\text{ radians} \approx 0.01746\text{ radians}$$If an arc of length $l$ of a circle of radius $r$ subtends a central angle $\theta$ measured in radians:
$$l = r\,\theta \iff \theta = \frac{l}{r}\quad(\theta\text{ MUST be in radians!})$$ $$\text{Area of Sector } A = \frac{1}{2}r^2\theta = \frac{1}{2}r\,l$$Convert $-47^\circ 30'$ into radian measure.
1. Convert minutes to degrees: $30' = \left(\dfrac{30}{60}\right)^\circ = \dfrac{1}{2}^\circ$.
2. Express total degrees: $-47^\circ 30' = -\left(47 + \dfrac{1}{2}\right)^\circ = -\left(\dfrac{95}{2}\right)^\circ$.
3. Multiply by $\dfrac{\pi}{180}$: $-\dfrac{95}{2} \times \dfrac{\pi}{180} = -\dfrac{19\pi}{72}\text{ radians}$.
A wheel makes $360$ revolutions in one minute. Through how many radians does it turn in one second?
1. Revolutions in $60\text{ seconds} = 360 \implies \text{Revolutions in } 1\text{ second} = \dfrac{360}{60} = 6$.
2. In $1$ complete revolution, the wheel turns through $2\pi\text{ radians}$.
3. In $6$ revolutions, total angle turned $= 6 \times 2\pi = \mathbf{12\pi\text{ radians}}$.
Find the degree measure of the angle subtended at the centre of a circle of radius $100\text{ cm}$ by an arc of length $22\text{ cm}$ (Use $\pi = \dfrac{22}{7}$).
1. Use $\theta = \dfrac{l}{r} = \dfrac{22}{100} = \dfrac{11}{50}\text{ radians}$.
2. Convert to degrees: $\theta = \dfrac{11}{50} \times \dfrac{180}{\pi} = \dfrac{11}{50} \times \dfrac{180 \times 7}{22} = \dfrac{126}{10}^\circ = 12.6^\circ$.
3. Express fractional part in minutes: $0.6^\circ = (0.6 \times 60)' = 36'$.
4. Therefore, $\theta = \mathbf{12^\circ 36'}$.
In Class 10, trigonometric ratios were defined only for acute angles in right triangles. In Class 11, we generalise trigonometry to Trigonometric Functions defined for all real numbers using the Unit Circle (circle of radius $r = 1$ centred at the origin $(0,0)$).
Let $P(x,y)$ be any point on the unit circle $x^2 + y^2 = 1$ such that the ray $OP$ makes an angle $t$ radians with the positive $x$-axis. Then:
$$\cos t = x \quad\text{and}\quad \sin t = y$$Since $(x,y)$ lies on $x^2 + y^2 = 1$, we immediately get the fundamental identity:
$$\cos^2 t + \sin^2 t = 1 \quad \forall t \in \mathbb{R}$$The other four functions are defined as quotients and reciprocals:
$$\tan t = \frac{\sin t}{\cos t} = \frac{y}{x}\ (x \neq 0),\quad \cot t = \frac{\cos t}{\sin t} = \frac{x}{y}\ (y \neq 0)$$ $$\sec t = \frac{1}{\cos t} = \frac{1}{x}\ (x \neq 0),\quad \csc t = \frac{1}{\sin t} = \frac{1}{y}\ (y \neq 0)$$| Function | Domain | Range | Fundamental Period |
|---|---|---|---|
| $y = \sin x$ | $\mathbb{R}$ | $[-1, 1]$ | $2\pi$ |
| $y = \cos x$ | $\mathbb{R}$ | $[-1, 1]$ | $2\pi$ |
| $y = \tan x$ | $\mathbb{R} \setminus \left\{(2n+1)\dfrac{\pi}{2}, n \in \mathbb{Z}\right\}$ | $\mathbb{R}$ | $\pi$ |
| $y = \cot x$ | $\mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$ | $\mathbb{R}$ | $\pi$ |
| $y = \sec x$ | $\mathbb{R} \setminus \left\{(2n+1)\dfrac{\pi}{2}, n \in \mathbb{Z}\right\}$ | $(-\infty, -1] \cup [1, \infty)$ | $2\pi$ |
| $y = \csc x$ | $\mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$ | $(-\infty, -1] \cup [1, \infty)$ | $2\pi$ |
A rigorous graphical understanding of trigonometric functions allows immediate visual verification of domain, range, periodicity, amplitude, intercepts, and asymptotic behavior.
If $\cos x = -\dfrac{3}{5}$, $x$ lies in the third quadrant, find the values of other five trigonometric functions.
1. In Quadrant III, only $\tan$ and $\cot$ are positive; $\sin, \cos, \sec, \csc$ are negative.
2. $\sin^2 x = 1 - \cos^2 x = 1 - \left(-\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25} \implies \sin x = -\sqrt{\dfrac{16}{25}} = \mathbf{-\dfrac{4}{5}}$.
3. $\tan x = \dfrac{\sin x}{\cos x} = \dfrac{-4/5}{-3/5} = \mathbf{\dfrac{4}{3}}$.
4. Reciprocals: $\cot x = \mathbf{\dfrac{3}{4}}$, $\sec x = \mathbf{-\dfrac{5}{3}}$, $\csc x = \mathbf{-\dfrac{5}{4}}$.
Find the value of $\sin\left(-\dfrac{11\pi}{3}\right)$ and $\cot\left(-\dfrac{15\pi}{4}\right)$.
1. For $\sin\left(-\dfrac{11\pi}{3}\right) = -\sin\left(\dfrac{11\pi}{3}\right) = -\sin\left(4\pi - \dfrac{\pi}{3}\right) = -\left(-\sin\dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{3} = \mathbf{\dfrac{\sqrt{3}}{2}}$.
2. For $\cot\left(-\dfrac{15\pi}{4}\right) = -\cot\left(\dfrac{15\pi}{4}\right) = -\cot\left(4\pi - \dfrac{\pi}{4}\right) = -\left(-\cot\dfrac{\pi}{4}\right) = \cot\dfrac{\pi}{4} = \mathbf{1}$.
Two angles are said to be allied if their sum or difference is either zero or an integral multiple of $\dfrac{\pi}{2}$ ($90^\circ$). The allied angles of $\theta$ include $(-\theta)$, $(\frac{\pi}{2} \pm \theta)$, $(\pi \pm \theta)$, $(\frac{3\pi}{2} \pm \theta)$, and $(2\pi \pm \theta)$.
To evaluate any expression of the form $T\left(n \cdot \dfrac{\pi}{2} \pm \theta\right)$:
| Angle $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
| $15^\circ$ $\left(\dfrac{\pi}{12}\right)$ | $\dfrac{\sqrt{6}-\sqrt{2}}{4} = \dfrac{\sqrt{3}-1}{2\sqrt{2}}$ | $\dfrac{\sqrt{6}+\sqrt{2}}{4} = \dfrac{\sqrt{3}+1}{2\sqrt{2}}$ | $2 - \sqrt{3}$ |
| $75^\circ$ $\left(\dfrac{5\pi}{12}\right)$ | $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ | $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ | $2 + \sqrt{3}$ |
| $18^\circ$ $\left(\dfrac{\pi}{10}\right)$ | $\dfrac{\sqrt{5}-1}{4}$ | $\dfrac{\sqrt{10+2\sqrt{5}}}{4}$ | $\sqrt{\dfrac{5-2\sqrt{5}}{5}}$ |
| $36^\circ$ $\left(\dfrac{\pi}{5}\right)$ | $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ | $\dfrac{\sqrt{5}+1}{4}$ | $\sqrt{5-2\sqrt{5}}$ |
| $22.5^\circ$ $\left(\dfrac{\pi}{8}\right)$ | $\dfrac{\sqrt{2-\sqrt{2}}}{2}$ | $\dfrac{\sqrt{2+\sqrt{2}}}{2}$ | $\sqrt{2} - 1$ |
Prove that: $\dfrac{\cos(\pi+\theta)\cos(-\theta)}{\sin(\pi-\theta)\cos\left(\dfrac{\pi}{2}+\theta\right)} = \cot^2\theta$.
1. Evaluate each term using allied angle rules:
2. Substitute into LHS: $\dfrac{(-\cos\theta)(\cos\theta)}{(\sin\theta)(-\sin\theta)} = \dfrac{-\cos^2\theta}{-\sin^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \mathbf{\cot^2\theta} = \text{RHS}$. $\blacksquare$
Find the exact value of $\tan 15^\circ$ and $\cos 105^\circ$.
1. $\tan 15^\circ = \tan(45^\circ - 30^\circ) = \dfrac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ\tan 30^\circ} = \dfrac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \dfrac{\sqrt{3}-1}{\sqrt{3}+1} = \mathbf{2 - \sqrt{3}}$.
2. $\cos 105^\circ = \cos(90^\circ + 15^\circ) = -\sin 15^\circ = -\left(\dfrac{\sqrt{6}-\sqrt{2}}{4}\right) = \mathbf{\dfrac{\sqrt{2}-\sqrt{6}}{4}}$.
An algebraic sum or difference of two or more angles is called a compound angle, such as $A+B$, $A-B$, or $A+B+C$.
Consider a unit circle centred at the origin $O$. Take four points on the circle:
Notice that $\angle P_1 O P_2 = y$ and $\angle P_0 O P_3 = y$. Thus, $\triangle P_1 O P_2 \cong \triangle P_3 O P_0$. Consequently, chord $P_1 P_3 = P_2 P_0$, which means $P_1 P_3^2 = P_2 P_0^2$.
Using the Euclidean distance formula:
$$P_2 P_0^2 = [\cos(x+y) - 1]^2 + [\sin(x+y) - 0]^2 = \cos^2(x+y) - 2\cos(x+y) + 1 + \sin^2(x+y) = 2 - 2\cos(x+y)$$ $$P_1 P_3^2 = [\cos x - \cos y]^2 + [\sin x - (-\sin y)]^2 = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y - \sin x \sin y)$$ $$= 2 - 2(\cos x \cos y - \sin x \sin y)$$Equating both expressions: $2 - 2\cos(x+y) = 2 - 2(\cos x \cos y - \sin x \sin y) \implies \mathbf{\cos(x+y) = \cos x \cos y - \sin x \sin y}$. $\blacksquare$
Prove that: $\dfrac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)} = \cot^2 x$.
$\text{LHS} = \dfrac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \dfrac{-\cos^2 x}{-\sin^2 x} = \cot^2 x = \text{RHS}$. $\blacksquare$
If $\tan A - \tan B = x$ and $\cot B - \cot A = y$, prove that $\cot(A - B) = \dfrac{1}{x} + \dfrac{1}{y}$.
1. Given: $x = \tan A - \tan B$, and $y = \cot B - \cot A = \dfrac{1}{\tan B} - \dfrac{1}{\tan A} = \dfrac{\tan A - \tan B}{\tan A \tan B} = \dfrac{x}{\tan A \tan B}$.
2. This gives: $\tan A \tan B = \dfrac{x}{y}$.
3. Now, $\cot(A - B) = \dfrac{1}{\tan(A - B)} = \dfrac{1 + \tan A \tan B}{\tan A - \tan B} = \dfrac{1 + x/y}{x} = \dfrac{y + x}{xy} = \mathbf{\dfrac{1}{x} + \dfrac{1}{y}}$. Hence Proved. $\blacksquare$
Prove that: $\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$.
1. Write $3x = 2x + x \implies \tan 3x = \tan(2x + x) = \dfrac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$.
2. Cross-multiply: $\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x$.
3. Expand: $\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x$.
4. Rearrange: $\mathbf{\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x}$. Hence Proved. $\blacksquare$
Transformation formulas allow us to convert products into sums/differences and sums/differences into products. They are indispensable for solving integrals, proving identities, and simplifying trigonometric equations.
Let $C = A+B$ and $D = A-B \implies A = \dfrac{C+D}{2}$, $B = \dfrac{C-D}{2}$:
$$\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$ $$\sin C - \sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$$ $$\cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$ $$\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right) = 2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)$$Prove that: $\dfrac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$.
1. Apply C-D formula to numerator: $\sin 5x + \sin 3x = 2\sin\left(\dfrac{5x+3x}{2}\right)\cos\left(\dfrac{5x-3x}{2}\right) = 2\sin 4x \cos x$.
2. Apply C-D formula to denominator: $\cos 5x + \cos 3x = 2\cos\left(\dfrac{5x+3x}{2}\right)\cos\left(\dfrac{5x-3x}{2}\right) = 2\cos 4x \cos x$.
3. $\text{LHS} = \dfrac{2\sin 4x \cos x}{2\cos 4x \cos x} = \dfrac{\sin 4x}{\cos 4x} = \mathbf{\tan 4x} = \text{RHS}$. $\blacksquare$
Prove that: $\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \dfrac{1}{16}$.
1. Substitute known value: $\cos 60^\circ = \dfrac{1}{2}$. Expression becomes: $\dfrac{1}{2} (\cos 20^\circ \cos 40^\circ \cos 80^\circ)$.
2. Group first pair and multiply/divide by $2$: $\dfrac{1}{4} [2\cos 40^\circ \cos 20^\circ] \cos 80^\circ$.
3. Use $2\cos A\cos B = \cos(A+B) + \cos(A-B)$: $= \dfrac{1}{4} [\cos 60^\circ + \cos 20^\circ] \cos 80^\circ = \dfrac{1}{4} \left[\dfrac{1}{2} + \cos 20^\circ\right] \cos 80^\circ$.
4. Expand: $= \dfrac{1}{8}\cos 80^\circ + \dfrac{1}{4}\cos 80^\circ \cos 20^\circ = \dfrac{1}{8}\cos 80^\circ + \dfrac{1}{8}[2\cos 80^\circ \cos 20^\circ]$.
5. Transform product: $= \dfrac{1}{8}\cos 80^\circ + \dfrac{1}{8}[\cos 100^\circ + \cos 60^\circ]$.
6. Note: $\cos 100^\circ = \cos(180^\circ - 80^\circ) = -\cos 80^\circ$. Thus: $\cos 80^\circ + (-\cos 80^\circ) = 0$.
7. Result: $\dfrac{1}{8}\cos 60^\circ = \dfrac{1}{8} \cdot \dfrac{1}{2} = \mathbf{\dfrac{1}{16}} = \text{RHS}$. $\blacksquare$
Formulas for $2A$, $3A$, and $\frac{A}{2}$ are direct consequences of compound angle addition rules where we set $y = x$.
Power Reduction Forms (Vital for Calculus & Integration):
$$1 + \cos 2A = 2\cos^2 A \iff \cos^2 A = \frac{1 + \cos 2A}{2}$$ $$1 - \cos 2A = 2\sin^2 A \iff \sin^2 A = \frac{1 - \cos 2A}{2}$$ $$\tan^2 A = \frac{1 - \cos 2A}{1 + \cos 2A}$$Prove that: $\cos 4x = 1 - 8\sin^2 x \cos^2 x$.
1. Write $\cos 4x = \cos(2(2x)) = 1 - 2\sin^2(2x)$.
2. Substitute $\sin 2x = 2\sin x \cos x$:
$\cos 4x = 1 - 2(2\sin x \cos x)^2 = 1 - 2(4\sin^2 x \cos^2 x) = \mathbf{1 - 8\sin^2 x \cos^2 x} = \text{RHS}$. $\blacksquare$
Find $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$, and $\tan\dfrac{x}{2}$ if $\tan x = -\dfrac{4}{3}$, and $x$ is in quadrant II.
1. Since $x \in (\pi/2, \pi)$, $\dfrac{x}{2} \in (\pi/4, \pi/2)$ (Quadrant I). Hence, all half-angle ratios are positive.
2. Given $\tan x = -4/3 \implies \sec^2 x = 1 + (-4/3)^2 = 25/9 \implies \cos x = -3/5$ (since $x$ in Q2).
3. $\sin\dfrac{x}{2} = \sqrt{\dfrac{1 - \cos x}{2}} = \sqrt{\dfrac{1 - (-3/5)}{2}} = \sqrt{\dfrac{8/5}{2}} = \sqrt{\dfrac{4}{5}} = \mathbf{\dfrac{2}{\sqrt{5}} = \dfrac{2\sqrt{5}}{5}}$.
4. $\cos\dfrac{x}{2} = \sqrt{\dfrac{1 + \cos x}{2}} = \sqrt{\dfrac{1 + (-3/5)}{2}} = \sqrt{\dfrac{2/5}{2}} = \sqrt{\dfrac{1}{5}} = \mathbf{\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}}$.
5. $\tan\dfrac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)} = \dfrac{2/\sqrt{5}}{1/\sqrt{5}} = \mathbf{2}$.
When angles are constrained by a relation, most notably in a triangle where $A + B + C = \pi$ ($180^\circ$), special cyclic identities emerge.
Major Standard Conditional Identities:
$$\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$$ $$\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$$ $$\sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$ $$\tan A + \tan B + \tan C = \tan A \tan B \tan C$$ $$\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1$$In any $\triangle ABC$, prove that: $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$.
1. Combine first two terms using C-D formula: $\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.
2. Since $A+B = \pi - C \implies \sin(A+B) = \sin C$:
$\sin 2A + \sin 2B = 2\sin C \cos(A-B)$.
3. Write $\sin 2C = 2\sin C \cos C$. The expression becomes:
$2\sin C [\cos(A-B) + \cos C]$.
4. Since $\cos C = \cos(\pi - (A+B)) = -\cos(A+B)$:
$2\sin C [\cos(A-B) - \cos(A+B)] = 2\sin C [2\sin A \sin B] = \mathbf{4\sin A \sin B \sin C} = \text{RHS}$. $\blacksquare$
If $A+B+C = \pi$, prove that: $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
1. $A + B = \pi - C \implies \tan(A+B) = \tan(\pi - C) = -\tan C$.
2. Expand LHS: $\dfrac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$.
3. Cross multiply: $\tan A + \tan B = -\tan C (1 - \tan A \tan B) = -\tan C + \tan A \tan B \tan C$.
4. Transpose $-\tan C$: $\mathbf{\tan A + \tan B + \tan C = \tan A \tan B \tan C}$. Hence Proved. $\blacksquare$
Equations involving trigonometric functions of unknown angles are called trigonometric equations.
Divide the entire equation by $\sqrt{a^2 + b^2}$. Let $\cos\phi = \dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi = \dfrac{b}{\sqrt{a^2+b^2}}$.
Then the equation reduces to $\cos(x - \phi) = \dfrac{c}{\sqrt{a^2+b^2}}$, which has real solutions if and only if $|c| \le \sqrt{a^2+b^2}$.
Find the principal and general solutions of the equation $\sin x = -\dfrac{\sqrt{3}}{2}$.
1. Recall $\sin(\pi/3) = \sqrt{3}/2$. Since $\sin x < 0$, $x$ lies in Quadrants III and IV.
2. In Q3: $x = \pi + \dfrac{\pi}{3} = \mathbf{\dfrac{4\pi}{3}}$. In Q4: $x = 2\pi - \dfrac{\pi}{3} = \mathbf{\dfrac{5\pi}{3}}$.
Principal Solutions: $\dfrac{4\pi}{3}, \dfrac{5\pi}{3}$.
3. For General Solution, take smallest positive angle $\alpha = \dfrac{4\pi}{3}$ (or $-\dfrac{\pi}{3}$):
$\mathbf{x = n\pi + (-1)^n \left(\dfrac{4\pi}{3}\right), \quad n \in \mathbb{Z}}$ (or $x = n\pi + (-1)^{n+1}\dfrac{\pi}{3}$).
Solve the equation: $\cos x + \sin x = \sqrt{2}$.
1. Here $a = 1, b = 1 \implies \sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2}$.
2. Divide both sides by $\sqrt{2}$:
$\dfrac{1}{\sqrt{2}}\cos x + \dfrac{1}{\sqrt{2}}\sin x = 1 \implies \cos x \cos\dfrac{\pi}{4} + \sin x \sin\dfrac{\pi}{4} = 1 \implies \cos\left(x - \dfrac{\pi}{4}\right) = 1$.
3. Since $\cos\theta = 1 \iff \theta = 2n\pi$, we have:
$x - \dfrac{\pi}{4} = 2n\pi \implies \mathbf{x = 2n\pi + \dfrac{\pi}{4}, \quad n \in \mathbb{Z}}$.
In any $\triangle ABC$, let the angles be denoted by $A, B, C$ and the lengths of the sides opposite to these angles by $a, b, c$ respectively. Let $R$ be the circumradius and $s = \dfrac{a+b+c}{2}$ be the semi-perimeter.
1. The Sine Rule: The sides of a triangle are proportional to the sines of the angles opposite to them.
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = k$$2. The Cosine Rule:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc} \iff a^2 = b^2 + c^2 - 2bc\cos A$$ $$\cos B = \frac{c^2 + a^2 - b^2}{2ca} \iff b^2 = c^2 + a^2 - 2ca\cos B$$ $$\cos C = \frac{a^2 + b^2 - c^2}{2ab} \iff c^2 = a^2 + b^2 - 2ab\cos C$$3. Projection Formulas:
$$a = b\cos C + c\cos B, \quad b = c\cos A + a\cos C, \quad c = a\cos B + b\cos A$$4. Napier's Analogy (Law of Tangents):
$$\tan\left(\frac{B-C}{2}\right) = \left(\frac{b-c}{b+c}\right)\cot\frac{A}{2}$$ $$\tan\left(\frac{C-A}{2}\right) = \left(\frac{c-a}{c+a}\right)\cot\frac{B}{2}, \quad \tan\left(\frac{A-B}{2}\right) = \left(\frac{a-b}{a+b}\right)\cot\frac{C}{2}$$In any $\triangle ABC$, prove that: $\dfrac{a - b}{c} = \dfrac{\sin\left(\dfrac{A-B}{2}\right)}{\cos\left(\dfrac{C}{2}\right)}$.
1. By Sine rule: $a = 2R\sin A$, $b = 2R\sin B$, $c = 2R\sin C$.
2. $\text{LHS} = \dfrac{2R\sin A - 2R\sin B}{2R\sin C} = \dfrac{\sin A - \sin B}{\sin C}$.
3. Apply C-D formula in numerator: $\sin A - \sin B = 2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$.
4. In denominator, write $\sin C = 2\sin\left(\dfrac{C}{2}\right)\cos\left(\dfrac{C}{2}\right)$.
5. Since $\dfrac{A+B}{2} = 90^\circ - \dfrac{C}{2} \implies \cos\left(\dfrac{A+B}{2}\right) = \sin\left(\dfrac{C}{2}\right)$.
6. Substitute: $\text{LHS} = \dfrac{2\sin(C/2)\sin\left(\dfrac{A-B}{2}\right)}{2\sin(C/2)\cos\left(\dfrac{C}{2}\right)} = \mathbf{\dfrac{\sin\left(\dfrac{A-B}{2}\right)}{\cos\left(\dfrac{C}{2}\right)}} = \text{RHS}$. $\blacksquare$
In any $\triangle ABC$, prove that: $a(b\cos C - c\cos B) = b^2 - c^2$.
1. Substitute Cosine rule values: $\cos C = \dfrac{a^2+b^2-c^2}{2ab}$ and $\cos B = \dfrac{c^2+a^2-b^2}{2ca}$.
2. $\text{LHS} = a \left[ b\left(\dfrac{a^2+b^2-c^2}{2ab}\right) - c\left(\dfrac{c^2+a^2-b^2}{2ca}\right) \right]$.
3. Cancel $b$ and $c$: $= a \left[ \dfrac{a^2+b^2-c^2}{2a} - \dfrac{c^2+a^2-b^2}{2a} \right] = \dfrac{a}{2a}[(a^2+b^2-c^2) - (c^2+a^2-b^2)]$.
4. Simplify: $= \dfrac{1}{2}[2b^2 - 2c^2] = \mathbf{b^2 - c^2} = \text{RHS}$. $\blacksquare$
For any linear combination of $\sin x$ and $\cos x$ with the same argument:
$$f(x) = a\sin x + b\cos x + c$$The maximum and minimum values are:
$$\text{Maximum Value} = c + \sqrt{a^2 + b^2}$$ $$\text{Minimum Value} = c - \sqrt{a^2 + b^2}$$ $$\text{Range of } (a\sin x + b\cos x) \text{ is } \left[-\sqrt{a^2+b^2}, \;\sqrt{a^2+b^2}\right]$$Find the maximum and minimum values of $7\cos x + 24\sin x + 5$.
1. Here $a = 24$, $b = 7$, $c = 5$.
2. $\sqrt{a^2+b^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$.
3. Range of $7\cos x + 24\sin x$ is $[-25, 25]$.
4. Therefore: $\text{Maximum} = 5 + 25 = \mathbf{30}$, and $\text{Minimum} = 5 - 25 = \mathbf{-20}$.
If $\sin x + \sin^2 x = 1$, find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$.
1. From $\sin x + \sin^2 x = 1 \implies \sin x = 1 - \sin^2 x = \cos^2 x$.
2. Squaring both sides: $\sin^2 x = \cos^4 x \implies 1 - \cos^2 x = \cos^4 x \implies \cos^4 x + \cos^2 x = 1$.
3. Notice that: $\cos^8 x + 2\cos^6 x + \cos^4 x = (\cos^4 x + \cos^2 x)^2$.
4. Substituting the result: $(1)^2 = \mathbf{1}$.
| Identity Type | Core Formulas |
|---|---|
| Radian & Sector | $l = r\theta$, $\text{Area} = \frac{1}{2}r^2\theta$, $\pi\text{ rad} = 180^\circ$, $1\text{ rad} \approx 57^\circ 16' 22''$ |
| Pythagorean | $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, $1 + \cot^2 x = \csc^2 x$ |
| Compound Angle | $\cos(x\pm y) = \cos x\cos y \mp \sin x\sin y$, $\sin(x\pm y) = \sin x\cos y \pm \cos x\sin y$ |
| Double Angle | $\sin 2x = 2\sin x\cos x = \dfrac{2\tan x}{1+\tan^2 x}$, $\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = \dfrac{1-\tan^2 x}{1+\tan^2 x}$ |
| Triple Angle | $\sin 3x = 3\sin x - 4\sin^3 x$, $\cos 3x = 4\cos^3 x - 3\cos x$, $\tan 3x = \dfrac{3\tan x - \tan^3 x}{1-3\tan^2 x}$ |
| Transformation | $2\sin A\cos B = \sin(A+B)+\sin(A-B)$, $\sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$ |
| General Solutions | $\sin x = \sin\alpha \implies x = n\pi+(-1)^n\alpha$, $\cos x = \cos\alpha \implies x = 2n\pi\pm\alpha$, $\tan x = \tan\alpha \implies x = n\pi+\alpha$ |
| Triangle Rules | $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, $\cos A = \dfrac{b^2+c^2-a^2}{2bc}$ |