Trigonometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Six Trigonometric Ratios: In right triangle with angle $\theta$:

• $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$
• $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
• $\tan\theta = \frac{\sin\theta}{\cos\theta}$

Reciprocals:

• $\csc\theta = 1/\sin\theta$
• $\sec\theta = 1/\cos\theta$
• $\cot\theta = 1/\tan\theta = \cos\theta/\sin\theta$

Fundamental Identities:

• $\sin^2\theta + \cos^2\theta = 1$
• $1 + \tan^2\theta = \sec^2\theta$
• $1 + \cot^2\theta = \csc^2\theta$

Key Values:

Angle	$\sin$	$\cos$	$\tan$
$0°$	0	1	0
$30°$	$1/2$	$\sqrt{3}/2$	$1/\sqrt{3}$
$45°$	$\sqrt{2}/2$	$\sqrt{2}/2$	1
$60°$	$\sqrt{3}/2$	$1/2$	$\sqrt{3}$
$90°$	1	0	$\infty$

⚡ NDA Tip: Know these values cold. The trigonometry section tests quick recall of exact values. Also remember $\sin 30° = 1/2$, not $1/3$.

⚡ Common Mistake: In right triangle context, the “opposite” and “adjacent” are relative to the specific angle $\theta$. Always identify which angle you’re working with.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Compound Angle Formulas:

• $\sin(A+B) = \sin A \cos B + \cos A \sin B$
• $\sin(A-B) = \sin A \cos B - \cos A \sin B$
• $\cos(A+B) = \cos A \cos B - \sin A \sin B$
• $\cos(A-B) = \cos A \cos B + \sin A \sin B$
• $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
• $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Double Angle:

• $\sin 2A = 2 \sin A \cos A$
• $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
• $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$

Radians vs Degrees: $180° = \pi$ radians. To convert: multiply degrees by $\pi/180$ to get radians.

Height and Distance: For angle of elevation $\theta$ from horizontal:

• Height = distance $\times \tan\theta$ (if angle measured from point at same level)
• More generally, use right triangle trigonometry

Worked Examples:

Example 1: Simplify $\frac{\sin(A+B) - \sin(A-B)}{\cos(A+B) + \cos(A-B)}$.

Using formulas: Numerator: $\sin A \cos B + \cos A \sin B - (\sin A \cos B - \cos A \sin B) = 2\cos A \sin B$. Denominator: $\cos A \cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B = 2\cos A \cos B$. So ratio $= \frac{2\cos A \sin B}{2\cos A \cos B} = \frac{\sin B}{\cos B} = \tan B$.

Example 2: If $\sin\theta = 3/5$, find $\cos\theta$ and $\tan\theta$.

Since $\sin^2\theta + \cos^2\theta = 1$: $\cos^2\theta = 1 - \sin^2\theta = 1 - 9/25 = 16/25$. $\cos\theta = \pm 4/5$. Assuming $\theta$ in first quadrant (common for right triangle problems): $\cos\theta = 4/5$. $\tan\theta = \sin\theta/\cos\theta = (3/5)/(4/5) = 3/4$.

Example 3 (NDA Pattern): From a point on ground 50m from base of a tower, the angle of elevation is $30°$. Find height of tower.

Let height $h$. $\tan 30° = h/50$. $1/\sqrt{3} = h/50$ → $h = 50/\sqrt{3} = \frac{50\sqrt{3}}{3} \approx 28.87$ m.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Multiple Angle Formulas:

• $\sin 3A = 3\sin A - 4\sin^3 A$
• $\cos 3A = 4\cos^3 A - 3\cos A$
• $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

Transformation Formulas:

• $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
• $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$
• $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
• $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

Inverse Trigonometric Functions:

• $\sin(\sin^{-1} x) = x$ for $x \in [-1,1]$
• $\sin^{-1}(\sin x) = x$ only for $x \in [-\pi/2, \pi/2]$
• Domain and range matters:
  • $\sin^{-1}: [-1,1] \to [-\pi/2, \pi/2]$
  • $\cos^{-1}: [-1,1] \to [0, \pi]$
  • $\tan^{-1}: \mathbb{R} \to (-\pi/2, \pi/2)$

Principal Values:

• $\sin^{-1} x \in [-\pi/2, \pi/2]$
• $\cos^{-1} x \in [0, \pi]$
• $\tan^{-1} x \in (-\pi/2, \pi/2)$

Trigonometric Equations:

General solutions:

• $\sin\theta = \sin\alpha$ → $\theta = n\pi + (-1)^n \alpha$
• $\cos\theta = \cos\alpha$ → $\theta = 2n\pi \pm \alpha$
• $\tan\theta = \tan\alpha$ → $\theta = n\pi + \alpha$

Advanced Problems:

Problem: Solve $\sin 3x = \cos 2x$.

$\sin 3x = \cos 2x = \sin(\pi/2 - 2x)$. So $3x = n\pi + (-1)^n (\pi/2 - 2x)$. This gives two cases based on $n$ even or odd.

For $n = 2m$ (even): $3x = 2m\pi + (\pi/2 - 2x)$ → $5x = 2m\pi + \pi/2$ → $x = \frac{2m\pi}{5} + \frac{\pi}{10}$.

For $n = 2m+1$ (odd): $3x = (2m+1)\pi - (\pi/2 - 2x) = (2m+1)\pi - \pi/2 + 2x$ → $x = (2m+1)\pi - \pi/2 = (2m+1)\pi/2 - \pi/2 = m\pi$.

So solutions: $x = m\pi$ or $x = \frac{2m\pi}{5} + \frac{\pi}{10}$.

Problem: Show that $\tan(\pi/3) + \tan(\pi/12) = \tan(\pi/4)$? No.

Actually $\pi/3 + \pi/12 = 5\pi/12$, not $\pi/4$. The identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ holds when $A+B+C = \pi$.

NDA Exam Pattern:

• NDA Mathematics: 120 questions, 300 marks
• Trigonometry: 10-15 questions
• Focus on solving triangles, heights and distances
• Inverse trig functions appear less frequently
• Compound angle formulas are commonly applied
• Know exact values for standard angles $0°, 30°, 45°, 60°, 90°$

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