Trigonometric Ratios Of Some Specific Angles

Trigonometric Ratios Of Some Specific Angles

The angles 0°, 30°, 45°, 60°, 90° are angles for which we have values of T.R.

θ	0°	30°	45°	60°	90°
Sin	0	1/2	1/√2	√3/2	1
Cos	1	√3/2	1/√2	1/2	0
Tan	0	1/√3	1	√3	∞
Cosec	∞	2	√2	2/√3	1
Sec	1	2/√3	√2	2	∞
Cot	∞	√3	1	1/√3	0

Trigonometric Ratios Of Some Specific Angles With Examples

Example 1:    Evaluate each of the following in the simplest form:

(i) sin 60º cos 30º + cos 60º sin 30º

(ii) sin 60º cos 45º + cos 60º sin 45º

Sol.      (i)  sin 60º cos 30º + cos 60º sin 30º

(=frac{sqrt{3}}{2}times frac{sqrt{3}}{2}+frac{1}{2}times frac{1}{2}=frac{3}{4}+frac{1}{4}=1)

(ii)  sin 60º cos 45º + cos 60º sin 45º

( =frac{sqrt{3}}{2}times frac{1}{sqrt{2}}+frac{1}{2}times frac{1}{sqrt{2}} )

( =frac{sqrt{3}}{2sqrt{2}}+frac{1}{2sqrt{2}}=frac{sqrt{3}+1}{2sqrt{2}} )

Example 2:     Evaluate the following expression:

(i) tan 60º cosec² 45º + sec² 60º tan 45º

(ii) 4cot² 45º – sec² 60º + sin² 60º + cos² 90º.

Sol.      (i)  tan 60º cosec² 45º + sec² 60º tan 45º

= tan 60º (cosec 45º)² + (sec 60º)² tan 45º

= √3 × (√2)² + (2)² × 1

= √3  × 2 + 4 = 4 + 2 √3

(ii)  4cot² 45º – sec² 60º + sin² 60º + cos² 90º

= 4(cot 45º)² – (sec 60º)² + (sin 60º)² + (cos 90º)²

= 4 × (1)² – (2)² + (√3/2)² + 0

= 4 – 4 + 3/4 + 0 = 3/4

[youtube https://www.youtube.com/watch?v=jNrizjTD4F8?feature=oembed]

Example 3:     Show that:

(i) 2(cos²45º + tan²60º) – 6(sin²45º – tan²30º) = 6

(ii) 2(cos⁴60º + sin⁴30º) – (tan²60º + cot²45º) + 3 sec²30º = 1/4

Sol.    (i)  2(cos²45º + tan²60º) – 6(sin²45º – tan²30º)

( =2left( right)}^{2}}+ right)-6left( right)}^{2}}- right)}^{2}} right) )

( =2left( frac{1}{2}+3 right)-6left( frac{1}{2}-frac{1}{3} right)=2left( frac{1+6}{2} right)-6left( frac{3-2}{6} right) )

( =2times frac{7}{2}-6times frac{1}{6}=7-1=6 )

(ii)  2(cos⁴60º + sin⁴30º) – (tan²60º + cot²45º) + 3 sec²30º

( =2left( + right)-left( + right)+3 right)}^{2}} )

( =2left( frac{1}{16}+frac{1}{16} right)left( 3+1 right)+3times frac{4}{3} )

( =2times ~frac{1}{8}-4+4=frac{1}{4} )

Example 4:     Find the value of x in each of the following :

(i) tan 3x = sin 45º cos 45º + sin 30º

(ii) cos x = cos 60º cos 30º + sin 60º sin 30º

Sol.    (i)  tan 3x = sin 45º cos 45º + sin 30º

( Rightarrow tantext{ }3x=frac{1}{sqrt{2}}times frac{1}{sqrt{2}}+frac{1}{2} )

( Rightarrow tantext{ }3x=frac{1}{2}+frac{1}{2}  )

⇒ tan 3x = 1

⇒ tan 3x = tan 45º

⇒ 3x = 45º    ⇒      x = 15º

(ii)  cos x = cos 60º cos 30º + sin 60º sin 30º

( Rightarrow costext{ }x=frac{1}{2}times frac{sqrt{3}}{2}+frac{sqrt{3}}{2}times frac{1}{2} )

( Rightarrow costext{ }x=frac{sqrt{3}}{4}+frac{sqrt{3}}{4} )

⇒ cos x = √3/2

⇒ cos x = cos 30º

⇒ x = 30º

Example 5:     If x = 30°, verify that

((i)tan 2x=frac{2tan x}{1-x}text{             }(ii)sin x=sqrt{frac{1-cos 2x}{2}})

Sol.(i)   When x = 30°, we have 2x = 60° .

∴  tan 2x = tan 60° = √3

( text{And, }frac{2tan x}{1-x}=frac{2tan 30{}^circ }{1-30{}^circ } )

( =frac{2times frac{1}{sqrt{3}}}{1- right)}^{2}}} )

( =frac{2/sqrt{3}}{1-frac{1}{3}}=frac{2/sqrt{3}}{2/3}=frac{2}{sqrt{3}}times frac{3}{2}=sqrt{3} )

( tan 2x=frac{2tan x}{1-x} )

(ii)        When x = 30°, we have 2x = 60°.

( sqrt{frac{1-cos 2x}{2}}=sqrt{frac{1-cos 60{}^circ }{2}} )

( sqrt{frac{1-frac{1}{2}}{2}}=sqrt{frac{1}{4}}=frac{1}{2} )

And, sinx = sin30° = 1/2

( sin x=frac{sqrt{1-cos 2x}}{2} )

Example 6:     Find the value of θ in each of the following :

(i) 2 sin 2θ = √3      (ii) 2 cos 3θ = 1

(i)  2 sin 2θ= √3

⇒ sin 2θ= √3/2

⇒ sin 2θ= sin 60°

⇒ 2θ= 60°   ⇒    θ = 30°

(ii)  2 cos 3θ = 1

⇒ cos 3θ = 1/2

⇒ cos 3θ = cos 60°

⇒ 3θ = 60°   ⇒   θ = 20°.

Example 7:    If θ is an acute angle and sin θ = cos θ, find the value of 2 tan²θ + sin²θ – 1.

Sol. sin θ = cos θ

( Rightarrow frac{sin theta }{cos theta }=frac{cos theta }{cos theta } )

[Dividing both sides by cos θ]

⇒ tanθ = 1

⇒ tanθ = tan45°  ⇒  θ= 45°

∴ 2 tan²θ + sin²θ – 1

= 2tan²45° + sin²45° – 1

( =2+ right)}^{2}}-1  )

( =2+frac{1}{2}-1=frac{5}{2}-1=frac{3}{2} )

Example 8:     An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side.

Sol.    Let ABC be an equilateral triangle inscribed in a circle of radius 6 cm. Let O be the centre of the circle.



Then, OA = OB = OC = 6 cm.

Let OD be perpendicular from O on side BC. Then, D is mid-point of BC and OB and OC are bisectors of ∠B and ∠C respectively.

∴ ∠OBD = 30°

In ∆OBD, right angled at D, we have

∠OBD = 30° and OB = 6 cm.

( cos angle OBD=frac{BD}{OB}Rightarrow cos =frac{BD}{6} )

( Rightarrow BD=6cos =6times frac{sqrt{3}}{2}=3sqrt{3}text{ } )

⇒ BC = 2 BD = 2(3√3 )cm = 6 √3 cm.

Example 9:     Using the formula, sin(A – B) = sinA cosB – cosA sinB, find the value of sin 15º.

Sol.    Let A = 45º and B = 30º. Then A – B = 15º. Putting A = 45º and B = 30º in the given formula, we get

sin(45º – 30º) = sin45º cos30º – cos45º sin30º

( or,sin (45{}^text{o}-30{}^text{o})=frac{1}{sqrt{2}}times frac{sqrt{3}}{2}-frac{1}{sqrt{2}}times frac{1}{2}  )

( =frac{sqrt{3}-1}{2sqrt{2}}Rightarrow sin 15{}^text{o}=frac{sqrt{3}-1}{2sqrt{2}} )

Example 10:     If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90° ; A > B, find A and B.

tan (A + B) = √3 = tan 60° & tan (A – B) = 1/√3 = tan 30°

A + B = 60° …….(1)

A – B = 30° …….(2)

2A = 90°  ⇒  A = 45°

adding (1) & (2)

A + B = 60

A – B = 30

Subtract equation (2) from (1)

A + B = 60

A – B = 30

2B = 30°

⇒ B = 15°. Ans.

Note: sin(A + B) = sin A cos B + cos A sin B

sin(A + B) ≠ sin A + sin B.