Trigonometric Ratios of Standard Angles

Trigonometric Ratios of Standard Angles

We have to remember the values of trigonometric ratios of the standard angles like 0°, 30°, 45°, 60°, 90° and so on. Those values can be derived and understood as follows:

Trigonometric ratios of 45°

Let OX be the initial line and trace out an ∠XOP = 45°. From Q, a point on OP, draw QN⊥OX. Then ∠NQO = 90° - 45° = 45°. Therefore it is an isosceles triangle, i.e. ON = NQ. Let ON = NQ = a. Now, by Pythagoras relation,

OQ² = ON² + NQ²

= a² + a²

= 2a²

∴ OQ = a√2

Now for right angled triangle ONQ with angle of reference ∠NOQ = 45°,

Sin45° = NQ/OQ = a/a√2 = 1/√2

Cosec45° = OQ/NQ = a√2/a = √2

Cos45° = ON/OQ = a/a√2 = 1/√2

Sec45° = OQ/ON = a√2/a = √2

tan45° = NQ/ON = a/a = 1

cot45° = ON/NQ = a/a = 1

Trigonometric ratios of 60° and 30°

Let ABC be a triangle in which AB = BC = CA and hence ∠A = ∠B = ∠C = 60°. From A draw AD⊥BC. Then ∠BAD = ∠CAD = 30° and BD = CD. Let AB = BC = CA = 2a. So, BD = CD = a.

Now, from right angled DABD,

AD² = AB² – BD²

= 4a² – a²

= 3a²

∴ AD = a√3

Now, in right angled DABD with angle of reference ∠ABD = 60°,

Sin60° = AD/AB = a√3/2a = √3/2

Cosec60° = AB/AD = 2a/a√3 = 2/√3

Cos60° = BD/AB = a/2a = 1/2

Sec60° = AB/BD = 2a/a = 2

tan60° = AD/BD = a√3/a = √3

cot60° = BD/AD = a/a√3 = 1/√3

And, with the angle of reference ∠BAD = 30°,

Sin30° = BD/AB = a/2a = 1/2

Cosec30° = AB/BD = 2a/a = 2

Cos30° = AD/AB = a√3/2a = √3/2

Sec30° = AB/AD = 2a/a√3 = 2/√3

tan30° = BD/AD = a/a√3 = 1/√3

cot30° = AD/BD = a√3/a = √3

Trigonometric ratios of 0°

Let OX be the initial line and trace out an ∠XOP = θ. Let Q be a point on OP. Draw QM⊥OX.

When θ is small enough, QM will be small enough.

And, when θ = 0°, QM = 0 and OQ = OM. Now, from right angled triangle MOQ with angle of reference θ = 0°,

Sin0° = QM/OQ = 0/OQ = 0

Cosec0° = OQ/QM = OQ/0 = ∞ (undefined)

Cos0° = OM/OQ = OM/OM = 1

Sec0° = OQ/OM = OM/OM = 1

tan0° = QM/OM = 0/OM = 0

cot0° = OM/QM = OM/0 = ∞ (undefined)

Trigonometric ratios of 90°

Let OX be the initial line and trace out an ∠XOP = θ. Let Q be a point on OP. Draw QM⊥OX.

When ∠MOQ = θ is close to 90°, OM will be close to 0.

And, when θ = 90°, OM = 0 and OQ = MQ. Now, from right angled triangle MOQ with angle of reference θ = 90°,

Sin90° = MQ/OQ = MQ/MQ = 1

Cosec90° = OQ/MQ = MQ/MQ = 1

Cos90° = OM/OQ = 0/OM = 0

Sec90° = OQ/OM = OQ/0 = ∞ (undefined)

tan90° = MQ/OM = MQ/0 = ∞ (undefined)

cot90° = OM/MQ = 0/MQ = 0

From the above relations, the values of various trigonometric ratios of standard angles can be summarized as follows:

Table of trigonometric ratios of standard angles: 0°, 30°, 45°, 60° and 90°.

Worked Out Examples

Example 1: Find the value of sin230° + sin245° + sin260° Solution: Here, sin230° + sin245° + sin260° = (1/2)^2+ (1/√2)^2+ (√3/2)^2 = 1/4+ 1/2+ 3/4 = (1 + 2 + 3)/4 = 6/4 = 3/2

Example 2: Find the value of tan 〖π/3〗^csin 〖π/3〗^c+ sin 〖π/4〗^ccos 〖π/2〗^c+ cos 〖π/2〗^csin 〖π/3〗^c Solution: Here, tan 〖π/3〗^csin 〖π/3〗^c+ sin 〖π/4〗^ccos 〖π/2〗^c+ cos 〖π/2〗^csin 〖π/3〗^c = tan60°sin60° + sin45°cos90° + cos90°sin60° = √3 × √3/2 + 1/√2 × 0 + 0 × √3/2 = 3/2 + 0 + 0 = 3/2

Example 3: Prove that: (1 + tan30°)/(1 - tan30°) = (cos30° + sin30°)/(cos30° - sin30°) Solution: Here, L.H.S. = (1 + tan30°)/(1 - tan30°) = (1 + 1/√3)/(1- 1/√3) = ( (√3 + 1)/√3)/((√3- 1)/√3) = (√3 + 1)/(√3- 1) R.H.S. = (cos30° + sin30°)/(cos30° - sin30°) = (√3/2 + 1/2)/(√3/2 + 1/2) = ( (√3 + 1)/√3)/((√3- 1)/√3) = (√3 + 1)/(√3- 1) ∴ L.H.S. = R.H.S. Proved.

Example 5: In the given figure, the distance between house B and House C is 20m and ∠BCA = 30°. Calculate the distance between house A and B.