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04
LESSON 04 · CLASS 10 MATHEMATICS

Standard Values

Learn the exact values of trigonometric ratios for 0°, 30°, 45°, 60°, and 90° with easy memory tricks, geometric proofs, and step-by-step examples.

01 · THE BASICS

What are Standard Values?

Standard values are fixed, exact numerical values of trigonometric ratios for five specific angles:0°, 30°, 45°, 60°, and 90°.

In Class 10 Mathematics, these five angles are fundamental because their trigonometric ratios can be derived using simple geometry without needing a calculator.

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Key Idea

Memorising standard values makes solving trigonometric expressions, heights and distances, and algebraic simplification much faster.

02 · VALUE TABLE

Trigonometric Values Table

Here is the complete reference table for all 6 trigonometric ratios across the standard angles:

Ratio (θ)
0°
30°
45°
60°
90°
sin θ
0
12
1√2
√32
1
cos θ
1
√32
1√2
12
0
tan θ
0
1√3
1
√3
Not Defined
cosec θ
Not Defined
2
√2
2√3
1
sec θ
1
2√3
√2
2
Not Defined
cot θ
Not Defined
√3
1
1√3
0
03 · MEMORY TRICKS

How to Remember the Table Easily

You do not need to memorize all 30 values! Follow these 3 simple tricks:

TRICK 1

The Square Root Pattern for Sin θ

Write numbers 0 to 4 under each angle, divide by 4, and take the square root:

0°√(0 / 4)= 0
30°√(1 / 4)= 1/2
45°√(2 / 4)= 1/√2
60°√(3 / 4)= √3/2
90°√(4 / 4)= 1
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Cos θ is Reverse of Sin θ

Just copy the sin θ values from right to left (90° down to 0°) to get all cos θ values.

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Tan θ = Sin θ ÷ Cos θ

Divide the top number of sin θ by the top number of cos θ to get tan θ instantly!

04 · GEOMETRIC PROOFS

Where do these values come from?

Let's derive the values for 45°, 30°, and 60° using simple right triangles:

PROOF 1

Trigonometric Ratios of 45°

45°aaa√2

Consider an isosceles right-angled triangle with two equal sides of length a:

  • Perpendicular = a
  • Base = a
  • By Pythagoras theorem: Hypotenuse = √(a² + a²) = a√2

sin 45° = Perpendicular / Hypotenuse = a / (a√2) = 1/√2

cos 45° = Base / Hypotenuse = a / (a√2) = 1/√2

tan 45° = Perpendicular / Base = a / a = 1

PROOF 2

Trigonometric Ratios of 30° and 60°

60°30°a√3a2a

Consider an equilateral triangle of side 2a split by an altitude:

  • Base = a
  • Hypotenuse = 2a
  • By Pythagoras theorem: Altitude = √((2a)² - a²) = a√3

sin 30° = Base / Hypotenuse = a / 2a = 1/2

sin 60° = Altitude / Hypotenuse = (a√3) / 2a = √3/2

tan 60° = Altitude / Base = (a√3) / a = √3

05 · SOLVED EXAMPLES

Practice Problems

EXAMPLE 1

Evaluate: sin 60° cos 30° + sin 30° cos 60°

Solution:

Substitute the standard values from the table:

  • sin 60° = √3/2
  • cos 30° = √3/2
  • sin 30° = 1/2
  • cos 60° = 1/2
Expression = (√3/2 × √3/2) + (1/2 × 1/2)
= (3/4) + (1/4)
= 4/4 = 1
EXAMPLE 2

Evaluate: 2 tan² 45° + cos² 30° - sin² 60°

Solution:

  • tan 45° = 1 → tan² 45° = 1² = 1
  • cos 30° = √3/2 → cos² 30° = 3/4
  • sin 60° = √3/2 → sin² 60° = 3/4
Expression = 2(1) + (3/4) - (3/4)
= 2 + 0 = 2
06 · SUMMARY

Quick Revision

01

Standard values apply to 0°, 30°, 45°, 60°, and 90°.

02

Use the pattern √(n)/2 for n = 0, 1, 2, 3, 4 to get sin values.

03

cos θ values are simply the reverse order of sin θ values.

04

tan 90°, cosec 0°, sec 90°, and cot 0° are Not Defined (division by 0).