The Standard Values Table

A handful of angles appear again and again: 0°, 30°, 45°, 60°, 90°. Their exact ratio values are worth memorising cold.

Ratio 0° 30° 45° 60° 90°
sin⁡\sin 0 12\tfrac12 12\tfrac{1}{\sqrt2} 32\tfrac{\sqrt3}{2} 1
cos⁡\cos 1 32\tfrac{\sqrt3}{2} 12\tfrac{1}{\sqrt2} 12\tfrac12 0
tan⁡\tan 0 13\tfrac{1}{\sqrt3} 1 3\sqrt3 not defined

Key Point: Notice cos⁡\cos is just sin⁡\sin read backwards. And tan⁡=sin⁡/cos⁡\tan = \sin/\cos.

[Board Important] tan⁡90°\tan 90° is not defined because cos⁡90°=0\cos 90° = 0 and we cannot divide by zero. Similarly sec⁡90°\sec 90° and csc⁡0°\csc 0° are undefined.

A Memory Trick for sin

There's a neat pattern. Write 0,1,2,3,40, 1, 2, 3, 4 under the angles 0°,30°,45°,60°,90°0°, 30°, 45°, 60°, 90°, then take   \sqrt{\;} and divide by 2:

sin⁡θ=04,14,24,34,44=0,12,12,32,1\sin\theta = \sqrt{\frac{0}{4}}, \sqrt{\frac{1}{4}}, \sqrt{\frac{2}{4}}, \sqrt{\frac{3}{4}}, \sqrt{\frac{4}{4}} = 0, \tfrac12, \tfrac{1}{\sqrt2}, \tfrac{\sqrt3}{2}, 1

Then cos⁡\cos is the same list reversed.

Key Point: sin⁡θ=n2\sin\theta = \dfrac{\sqrt{n}}{2} where n=0,1,2,3,4n = 0,1,2,3,4 for 0°,30°,45°,60°,90°0°,30°,45°,60°,90°.

[JEE/NEET Tip] This n/2\sqrt{n}/2 trick is the fastest way to reconstruct the table if you blank out in an exam.

Where the Values Come From

The values aren't magic — they come from two special triangles:

  • A 45°-45°-90° triangle has legs equal, so if each leg is 1, the hypotenuse is 2\sqrt2, giving sin⁡45°=cos⁡45°=12\sin 45° = \cos 45° = \tfrac{1}{\sqrt2}.
  • A 30°-60°-90° triangle (half an equilateral triangle of side 2) has sides 1,3,21, \sqrt3, 2, giving sin⁡30°=12\sin 30° = \tfrac12, sin⁡60°=32\sin 60° = \tfrac{\sqrt3}{2}, etc.

Key Point: Remember the two special triangles and you can re-derive every standard value.

[Board Important] Drawing the special triangle is a valid method to justify a value if you forget the table.

Reciprocal Ratios at Standard Angles

From the table, the reciprocal ratios follow at once:

  • csc⁡30°=2\csc 30° = 2, csc⁡45°=2\csc 45° = \sqrt2, csc⁡60°=23\csc 60° = \tfrac{2}{\sqrt3}.
  • sec⁡30°=23\sec 30° = \tfrac{2}{\sqrt3}, sec⁡45°=2\sec 45° = \sqrt2, sec⁡60°=2\sec 60° = 2.
  • cot⁡30°=3\cot 30° = \sqrt3, cot⁡45°=1\cot 45° = 1, cot⁡60°=13\cot 60° = \tfrac{1}{\sqrt3}.

Key Point: cosec, sec, cot at standard angles are just reciprocals of sin, cos, tan.

[Board Important] sec⁡0°=1\sec 0° = 1 and csc⁡90°=1\csc 90° = 1; but sec⁡90°\sec 90° and csc⁡0°\csc 0° are undefined (division by zero).

Solved Examples

Example 1: Direct evaluation

Evaluate sin⁡30°+cos⁡60°\sin 30° + \cos 60°.

Solution:

  1. sin⁡30°=12\sin 30° = \tfrac12, cos⁡60°=12\cos 60° = \tfrac12.
  2. Sum =12+12=1= \tfrac12 + \tfrac12 = 1.

Final Answer: 1.

Takeaway: Read straight off the table.

Example 2: Product

Evaluate sin⁡60°cos⁡30°+cos⁡60°sin⁡30°\sin 60° \cos 30° + \cos 60° \sin 30°.

Solution:

  1. =32⋅32+12⋅12=34+14=1= \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} + \tfrac12\cdot\tfrac12 = \tfrac34 + \tfrac14 = 1.

Final Answer: 1.

Takeaway: This is sin⁡(60°+30°)=sin⁡90°=1\sin(60°+30°) = \sin 90° = 1 — a nice check.

Example 3: tan values

Evaluate tan⁡260°+tan⁡245°\tan^2 60° + \tan^2 45°.

Solution:

  1. tan⁡60°=3\tan 60° = \sqrt3, so tan⁡260°=3\tan^2 60° = 3.
  2. tan⁡45°=1\tan 45° = 1, so tan⁡245°=1\tan^2 45° = 1.
  3. Sum =3+1=4= 3 + 1 = 4.

Final Answer: 4.

Takeaway: Square after substituting the value.

Example 4: Mixed expression

Evaluate tan⁡30°cot⁡60°\dfrac{\tan 30°}{\cot 60°}.

Solution:

  1. tan⁡30°=13\tan 30° = \tfrac{1}{\sqrt3}, cot⁡60°=13\cot 60° = \tfrac{1}{\sqrt3}.
  2. Ratio =1/31/3=1= \dfrac{1/\sqrt3}{1/\sqrt3} = 1.

Final Answer: 1.

Takeaway: tan⁡30°=cot⁡60°\tan 30° = \cot 60° — a complementary-angle hint.

Example 5: Find the angle

If sin⁡θ=32\sin\theta = \tfrac{\sqrt3}{2} and θ\theta is acute, find θ\theta.

Solution:

  1. From the table, sin⁡60°=32\sin 60° = \tfrac{\sqrt3}{2}.

Final Answer: θ=60°\theta = 60°.

Takeaway: Reading the table backwards finds the angle.

Example 6: Evaluate with squares

Evaluate 4sin⁡260°−3tan⁡230°4\sin^2 60° - 3\tan^2 30°.

Solution:

  1. sin⁡260°=34\sin^2 60° = \tfrac34, so 4⋅34=34\cdot\tfrac34 = 3.
  2. tan⁡230°=13\tan^2 30° = \tfrac13, so 3⋅13=13\cdot\tfrac13 = 1.
  3. 3−1=23 - 1 = 2.

Final Answer: 2.

Takeaway: Square the value first, then multiply by the coefficient.

Example 7: Fraction

Evaluate cos⁡45°sec⁡30°+csc⁡30°\dfrac{\cos 45°}{\sec 30° + \csc 30°}.

Solution:

  1. cos⁡45°=12\cos 45° = \tfrac{1}{\sqrt2}; sec⁡30°=23\sec 30° = \tfrac{2}{\sqrt3}; csc⁡30°=2\csc 30° = 2.
  2. Denominator =23+2=2+233= \tfrac{2}{\sqrt3} + 2 = \tfrac{2 + 2\sqrt3}{\sqrt3}.
  3. 1/2(2+23)/3=32 (2+23)=322(1+3)\dfrac{1/\sqrt2}{(2+2\sqrt3)/\sqrt3} = \dfrac{\sqrt3}{\sqrt2 \,(2 + 2\sqrt3)} = \dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)}.
  4. Rationalise: =3(3−1)22(3−1)=3−342= \dfrac{\sqrt3(\sqrt3 - 1)}{2\sqrt2(3-1)} = \dfrac{3 - \sqrt3}{4\sqrt2}.

Final Answer: 3−342\dfrac{3 - \sqrt3}{4\sqrt2}.

Takeaway: Substitute, combine, then rationalise the surd.

Example 8: Verify an identity numerically

Verify sin⁡230°+cos⁡230°=1\sin^2 30° + \cos^2 30° = 1.

Solution:

  1. sin⁡230°=14\sin^2 30° = \tfrac14, cos⁡230°=34\cos^2 30° = \tfrac34.
  2. Sum =14+34=1= \tfrac14 + \tfrac34 = 1.

Final Answer: 1 — verified.

Takeaway: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 holds for every angle.

Example 9: Solve for angle in an equation

Find acute θ\theta if 2cos⁡θ=12\cos\theta = 1.

Solution:

  1. cos⁡θ=12\cos\theta = \tfrac12.
  2. From the table, cos⁡60°=12\cos 60° = \tfrac12.

Final Answer: θ=60°\theta = 60°.

Takeaway: Isolate the ratio, then read the angle.

Example 10: Combine angles

Evaluate sin⁡90°−2cos⁡245°+tan⁡260°\sin 90° - 2\cos^2 45° + \tan^2 60°.

Solution:

  1. sin⁡90°=1\sin 90° = 1.
  2. cos⁡245°=12\cos^2 45° = \tfrac12, so 2⋅12=12\cdot\tfrac12 = 1.
  3. tan⁡260°=3\tan^2 60° = 3.
  4. 1−1+3=31 - 1 + 3 = 3.

Final Answer: 3.

Takeaway: Evaluate each term separately, then combine.