Important Special Values — The Extended Trig Table

"Master the rare angles to solve in seconds what others derive in minutes."

Introduction

Beyond the standard 0°, 30°, 45°, 60°, 90° lie angles like 15°, 18°, 22.5°, 36°, 54°, 72°, 75° that frequently appear in JEE problems. Deriving their values on the spot consumes precious time and mental energy. This article provides a systematic, memorable framework for these high-yield special values, turning complex evaluations into instant recall.

The Strategic Value Table: Grouped by Family

Grouping these angles by their mathematical relationships makes memorization logical, not just rote.

Family 1: The 15°–75° Pair (From 45° ± 30°)

These are the most frequently tested non-standard angles.

Angle (θ)	sin θ	cos θ	tan θ	Key Insight
15° (π/12)	$\displaystyle \frac{\sqrt{6} - \sqrt{2}}{4}$	$\displaystyle \frac{\sqrt{6} + \sqrt{2}}{4}$	$2 - \sqrt{3}$	$\theta = 45^\circ - 30^\circ$
75° (5π/12)	$\displaystyle \frac{\sqrt{6} + \sqrt{2}}{4}$	$\displaystyle \frac{\sqrt{6} - \sqrt{2}}{4}$	$2 + \sqrt{3}$	$\theta = 45^\circ + 30^\circ$

Simplified Forms (Easier to Recall & Use): $\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}, \quad \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}$ $\sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}}, \quad \cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

Memory Hacks for 15°/75°:

Sine 15° is Smaller: The numerator has "√3 minus 1".
Cosine 15° is Larger: The numerator has "√3 plus 1".
Complements: $\sin 75^\circ = \cos 15^\circ$ and $\cos 75^\circ = \sin 15^\circ$ .
Tangent: $\tan 15^\circ = 2 - \sqrt{3} \ (\approx 0.27)$ is small; $\tan 75^\circ = 2 + \sqrt{3} \ (\approx 3.73)$ is large.

Family 2: The 18°–36°–54°–72° Family (Pentagon Angles)

These angles are related to the regular pentagon (72°) and decagon (36°), and involve the Golden Ratio, $\phi = \frac{1+\sqrt{5}}{2} \approx 1.618$ .

Angle (θ)	sin θ	cos θ	tan θ	Note
18° (π/10)	$\displaystyle \frac{\sqrt{5} - 1}{4}$	$\displaystyle \frac{\sqrt{10 + 2\sqrt{5}}}{4}$	$\displaystyle \sqrt{\frac{5 - 2\sqrt{5}}{5}}$	Derived from pentagon geometry
36° (π/5)	$\displaystyle \frac{\sqrt{10 - 2\sqrt{5}}}{4}$	$\displaystyle \frac{\sqrt{5} + 1}{4} = \frac{\phi}{2}$	$\displaystyle \sqrt{5 - 2\sqrt{5}}$	Complementary to 54°
54° (3π/10)	$\displaystyle \frac{\sqrt{5} + 1}{4}$	$\displaystyle \frac{\sqrt{10 - 2\sqrt{5}}}{4}$	$\displaystyle \sqrt{\frac{5 + 2\sqrt{5}}{5}}$	$\sin 54^\circ = \cos 36^\circ$
72° (2π/5)	$\displaystyle \frac{\sqrt{10 + 2\sqrt{5}}}{4}$	$\displaystyle \frac{\sqrt{5} - 1}{4}$	$\displaystyle \sqrt{5 + 2\sqrt{5}}$	$\sin 72^\circ = \cos 18^\circ$

Key Symmetries to Remember:

$\sin 18^\circ = \cos 72^\circ = \frac{\sqrt{5}-1}{4}$
$\sin 36^\circ = \cos 54^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$
$\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$ (The Golden Ratio connection!)
$\sin 72^\circ = \cos 18^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$

Golden Ratio Insight: $\cos 36^\circ = \frac{\phi}{2}$ and $\sin 18^\circ = \frac{1}{2\phi}$ . This is a powerful memory hook.

Family 3: The 22.5°–67.5° Pair (Half of 45°)

Angle (θ)	sin θ	cos θ	tan θ	Derivation
22.5° (π/8)	$\displaystyle \frac{\sqrt{2 - \sqrt{2}}}{2}$	$\displaystyle \frac{\sqrt{2 + \sqrt{2}}}{2}$	$\sqrt{2} - 1$	Half-angle of 45°
67.5° (3π/8)	$\displaystyle \frac{\sqrt{2 + \sqrt{2}}}{2}$	$\displaystyle \frac{\sqrt{2 - \sqrt{2}}}{2}$	$\sqrt{2} + 1$	Complementary to 22.5°

Note: $\tan 22.5^\circ = \sqrt{2} - 1$ and $\tan 67.5^\circ = \sqrt{2} + 1$ are especially neat and frequently appear in problems.

Strategic Problem-Solving with Special Values

Problem Type 1: Direct Evaluation (JEE Main 2020 Style)

Question: If $\cos \theta = \frac{\sqrt{3} + 1}{2\sqrt{2}}$ and $\theta$ is acute, find $\theta$ .

Solution (Instant): Recognize the form as $\cos 15^\circ$ . Answer: $\theta = 15^\circ$ .

Problem Type 2: Expression Simplification

Question: Evaluate $\sin 75^\circ + \cos 75^\circ$ .

Solution (Fast Path): $\sin 75^\circ + \cos 75^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}} + \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{2\sqrt{3}}{2\sqrt{2}} = \sqrt{\frac{3}{2}}$ Time saved: ~90 seconds vs. deriving from compound angles.

Problem Type 3: Product Evaluation

Question: Find $\sin 15^\circ \cdot \sin 75^\circ$ .

Solution (Using memorized forms): $\sin 15^\circ \cdot \sin 75^\circ = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) = \frac{(\sqrt{3})^2 - 1^2}{8} = \frac{2}{8} = \frac{1}{4}$ Additional Insight: This also equals $\frac{1}{2} \sin 30^\circ = \frac{1}{4}$ , confirming the answer.

Applied Mastery: Breaking Down JEE Problems

📚 JEE Main 2022: Rational Expression with 18° and 36°

Question: Value of $\dfrac{3\cos 36^\circ + 5\sin 18^\circ}{5\cos 36^\circ - 3\sin 18^\circ}$ .

Strategic Solution:

Substitute known values:
- $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$
- $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$
Numerator: $3\cdot\frac{\sqrt{5}+1}{4} + 5\cdot\frac{\sqrt{5}-1}{4} = \frac{8\sqrt{5} - 2}{4} = 2\sqrt{5} - \frac{1}{2}$ .
Denominator: $5\cdot\frac{\sqrt{5}+1}{4} - 3\cdot\frac{\sqrt{5}-1}{4} = \frac{2\sqrt{5} + 8}{4} = \frac{\sqrt{5}}{2} + 2$ .
Simplify the ratio: After rationalization, this yields $\frac{17\sqrt{5} - 24}{11}$ .

Takeaway: Knowing the exact values allows direct substitution, bypassing complex trigonometric manipulations.

📚 JEE Main 2019: Polynomial Root Identification

Question: If $\alpha = \sin 36^\circ$ , it is a root of which equation?

Solution:

Recall: $\sin 36^\circ = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ .
Let $x = \sin 36^\circ$ . Then $16x^2 = 10 - 2\sqrt{5}$ . Isolate $\sqrt{5}$ and square again to eliminate the radical.
Process: $16x^2 - 10 = -2\sqrt{5}$ → $(16x^2 - 10)^2 = 20$ → $256x^4 - 320x^2 + 100 = 20$ .
Final Polynomial: $256x^4 - 320x^2 + 80 = 0$ → Divide by 16: $16x^4 - 20x^2 + 5 = 0$ .

Answer: Option (c). Recognizing the starting value was 90% of the solution.

Memory Consolidation: Patterns & Relationships

The "√3 ± 1" Pattern (15° & 75°)

Sine of the smaller angle (15°) has the minus sign.
Cosine of the smaller angle (15°) has the plus sign.
They swap for the complement (75°).

The Golden Ratio Pattern (18°, 36°, 54°, 72°)

Cos 36° and Sin 54° contain $\sqrt{5} + 1$ (the Golden Ratio φ).
Sin 18° and Cos 72° contain $\sqrt{5} - 1$ (the conjugate 1/φ).
The other sine/cosine values contain $\sqrt{10 \pm 2\sqrt{5}}$ .

The "√2 ± 1" Pattern (22.5° & 67.5°)

tan 22.5° = √2 - 1 (less than 1)
tan 67.5° = √2 + 1 (greater than 1)
Their product is $(\sqrt{2}-1)(\sqrt{2}+1) = 1$ , as expected for complementary tangents.

Quick Reference Decision Tree

Essential Derived Products & Sums

Memorizing these results can save additional steps.

Expression	Value	Why/Note
$\sin 15^\circ \cdot \sin 75^\circ$	$\frac{1}{4}$	= $\frac{1}{2} \cos 60^\circ$
$\cos 15^\circ \cdot \cos 75^\circ$	$\frac{1}{4}$	= $\frac{1}{2} \sin 30^\circ$
$\tan 15^\circ \cdot \tan 75^\circ$	$1$	Complementary angles
$\tan 15^\circ + \tan 75^\circ$	$4$	= $\frac{2}{\sin 30^\circ}$
$\sin 18^\circ \cdot \cos 36^\circ$	$\frac{1}{4}$	Interesting product
$\sin 36^\circ \cdot \cos 18^\circ$	$\frac{\sqrt{10-2\sqrt{5}} \cdot \sqrt{10+2\sqrt{5}}}{16} = \frac{\sqrt{80}}{16} = \frac{\sqrt{5}}{4}$
$\cos 36^\circ - \cos 72^\circ$	$\frac{1}{2}$	Common simplification

Practice for Fluency (Time Target: 2 min each)

Evaluate: $\sin^2 18^\circ + \cos^2 36^\circ$ .
Simplify: $\dfrac{\tan 22.5^\circ}{1 - \tan^2 22.5^\circ}$ . (Hint: This is $\frac{1}{2} \tan 45^\circ$ ?)
Find the value: $4 \sin 54^\circ \sin 66^\circ \sin 78^\circ$ . (Hint: 66° = 60°+6°, 78°=60°+18°; use product formulas from Article 3).
Prove: $\cos 36^\circ - \cos 72^\circ = \frac{1}{2}$ . (Use known values directly).
Solve: $\cos \theta = \frac{\sqrt{5} + 1}{4}$ for acute $\theta$ . What is $\tan \theta$ ?

Final Summary & Exam Strategy

Prioritize Memorization: The 15°/75° and 22.5°/67.5° families are highest yield. Know them perfectly.
Understand the Golden Ratio Link: You don't need to derive 18°/36° values if you remember $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$ . The others often follow by complement.
Use in Tandem with Other Tricks: Special values make the Substitution Method (Article 1) and Product Series (Article 3) even more powerful.
Verification: If you forget a value mid-exam, you can quickly derive it using:
- 15° = 45° - 30° (use compound angle formula).
- 22.5° = 45°/2 (use half-angle formula).
- 18° derivation is complex; this is the one set truly worth memorizing.
Time Investment: Spending 30 minutes to solidly memorize these values can save 10-15 minutes per paper—an excellent return on investment.

Integrating these special values into your toolkit makes you faster and more confident, turning intimidating problems into familiar friends.

"In the race against time, recognizing a special value is like finding a secret passage—it gets you to the answer while others are still mapping the maze."

Important Special Values