Trigonometry Exercise 13.2 – Questions:
I. Answer the following questions:
- What trigonometric ratios of angles from 0 to 90 are equal to 0?
- Which trigonometric ratios of angles from 0 to 90 are equal to 1?
- Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?
- Which trigonometric ratios of angles from 0 to 90 are not defined?
- Which trigonometric ratios of angles from 0 to 90 are equal?
II. Find θ. if 0≤θ≤90
- √2 cos θ = 1
- √3 tan θ = 1
- 2 sin θ = √3
- 5 sin θ = 0
- 3 tan θ = √3
III. Find the value of the following.
- sin 30˚ cos 60˚ – tan245˚
- sin 60˚ cos 30˚ + cos 60˚ sin 30˚
- cos 60˚ cos 30˚ – sin 60˚ sin 30˚
- 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚
- 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚
- cos 45˚/sec 30˚ + cosec 30˚
- (4sin260 – cos245)/(tan230 + sin20)
- (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)
- (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)
- (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)
IV. Prove the following equalities.
- sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
- 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
- if θ = 30˚, prove that 4cos2θ – 3 cos θ = cos 3θ
- If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3
- If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1
- If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB
Trigonometry Exercise 13.2 – Solutions:
I. Answer the following questions:
- What trigonometric ratios of angles from 0 to 90 are equal to 0?
Solution:
sinθ = 0 when θ = 0˚
cosθ = 0 when θ = 90˚
tanθ = 0 when θ = 0˚
cotθ = 0 when θ = 90˚
- Which trigonometric ratios of angles from 0 to 90 are equal to 1?
Solution:
sinθ = 1 when θ = 90˚
cosθ = 1 when θ = 0˚
tanθ = 1 when θ = 45˚
cosec θ = 1 when θ = 90 ˚
sec θ = 1 when θ = 0 ˚
- Which trigonometric ratios of angles from 0 to 90 are equal to 1/2?
Solution:
sinθ = 1/2 when θ = 30˚
cosθ = 1/2 when θ = 60˚
- Which trigonometric ratios of angles from 0 to 90 are not defined?
Solution:
tanθ = undefined when θ = 90˚
cosecθ = undefined when θ = 0˚
secθ = undefined when θ = 90˚
cotθ = undefined when θ = 0˚
- Find θ. if 0≤θ≤90
- √2 cos θ = 1
Solution:
√2 cos θ = 1
cos θ = 1/√2
We know, cos 45˚ = 1/√2
Therefore, cos θ = cos 45˚
Thus, θ = 45˚
- √3 tan θ = 1
Solution:
√3 tan θ = 1
tan θ = 1/√3
We know, tan 30˚ = 1/√3
Therefore, cos 30˚ = cos θ
Thus, θ = 30˚
- 2 sin θ = √3
Solution:
2 sin θ = √3
sin θ = √3/2
We know, sin 60˚ = √3/2
Therefore, sin 60˚ = sin θ
Thus, θ = 60˚
- 5 sin θ = 0
Solution:
5 sin θ = 0
sin θ = 0/5
sin θ = 0
We know, sin 0˚ = 0
Therefore, sin 0˚ = sin θ
Thus, θ = 0˚
- 3 tan θ = √3
Solution:
3 tan θ = √3
tan θ = √3/3
i.e., tan θ = 1/√3
We know, tan 30˚ = 1/√3
Therefore, tan 30˚ = tan θ
Thus, θ = 30˚
III. Find the value of the following.
- sin 30˚ cos 60˚ – tan245˚
Solution:
sin 30˚ cos 60˚ – tan245˚
= 1/2 x 1/2 – (1)2
= 1/4 – 1
= -3/4
- sin 60˚ cos 30˚ + cos 60˚ sin 30˚
Solution:
sin 60˚ cos 30˚ + cos 60˚ sin 30˚
= √3/2 x √3/2 + 1/2 x 1/2
= 3/4 + 1/4
= 3+1/4
= 4/4
= 1
- cos 60˚ cos 30˚ – sin 60˚ sin 30˚
Solution:
cos 60˚ cos 30˚ – sin 60˚ sin 30˚
= 1/2 x √3/2 – √3/2 x 1/2
= √3/4 – √3/2
= 0
- 2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚
Solution:
2sin230˚ – 3 cos2 30˚ + tan 60˚ + 3 sin2 90˚
= 2(1/2)2 – 3(√3/2 )2 + (√3) + 3(1)2
= 2(1/4) – 3(3/4) + √3 + 3
= 1/2 – 9/4 + √3 + 3
= 5/4 + √3
- 4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚
Solution:
4 sin2 60˚ + 3 tan2 30˚ – 8 sin 45˚ .cos 45˚
= 4(√3/2)2 + 3(1/√3)2 – 8(1/√2) .( 1/√2)
= 4(3/4) + 3(1/3) – 8(1/2)
= 3 + 1 – 4
= 0
- cos 45˚/(sec 30˚ + cosec 30˚)
Solution:
cos 45˚/(sec 30˚ + cosec 30˚)
= (1/√2)/( 2/√3+ 2)
= (1/√2)/(2+2√3/√3)
= √3/2√2(1+√3)
7.
- (sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)
Solution:
(sin30˚ + tan45˚ – cosec60˚)/(sec30˚ + cos 60˚ +cot45˚)
=(1/2 +1 – 2/√3)/(2/√3 + 1/2 + 1)
= (9 – 4√3/6)/(9 + 4√3/6)
= 9 – 4√3/9 – 4√3
- (5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)
Solution:
(5cos260˚ + 4 sec230˚ – tan245˚)/(sin230˚ + cos230˚)
= [5(1/2)2 + 4(2/√3)2 – (1)2]/[(1/2)2 + (√3/2)2]
= [5/4 + 16/3 – 1]/[1/4 + 3/4]
= (67/12)/(1)
= 67/12
- (5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)
Solution:
(5sin230˚ + cos245˚ – 4 tan230˚)/(2sin30˚ + cos30˚ + tan45˚)
= [5(1/2)2 + (1/√2)2 – 4(1/√3)2]/[2(1/2) + (√3/2) + 1]
= [5/4 + 1/2 – 4/3]/[1 + √3/2 + 1]
= [5/12]/[4+√3/2]
= 5/6(4+√3)
IV.Prove the following equalities.
- sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
Solution:
sin30˚ .cos60 ˚+ cos30˚.sin60˚ = sin 90˚
LHS = sin30˚ .cos60 ˚+ cos30˚.sin60˚
= 1/2 . 1/2 + √3/2.√3/2
= 1/4 + 3/4
= 1+3/4
= 1
RHS = sin 90˚ = 1
Therefore, LHS = RHS
- 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
Solution:
2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
Take, 2cos230˚ – 1 ………………..(1)
= 2(√3/2)2 – 1
= 2(3/4) – 1
= 3/2 – 1
= 3 – 2/2 = 1/2
1 – 2 sin2 30˚ ………………(2)
= 1 – 2(1/2)2
= 1 – 2(1/4)
= 1 – 1/2
= 1/2
cos60˚ …………………(3)
we know, cos60˚ = 1/2
Therefore, (1) = (2) = (3)
Thus, 2cos230˚ – 1 = 1 – 2sin230˚ = cos 60˚
- If θ = 30˚, prove that 4cos3θ – 3 cos θ = cos 3θ
Solution:
If θ = 30˚, we have to prove that 4cos3θ – 3 cos θ = cos 3θ
LHS = 4cos3θ – 3 cos θ
= 4cos330˚ – 3 cos 30˚
= 4(√3/2)3 – 3(√3/2)
= 4(3√3/8) – 3(√3/2)
= 3√3/2 – 3√3/2
= 3√3 – 3√3/2
= 0
RHS = cos 3θ
= cos 3(30˚)
= cos(90˚)
= 0
LHS = RHS
- If π = 180˚ and A = π/6 prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3
Solution:
If π = 180˚ and A = π/6 = 30˚ we have to prove that (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA) = 1/3
LHS = (1 – cosA)(1 + cos A)/(1 – sin A)(1+ sinA)
= (1 – cos 30˚)(1 + cos 30˚)/(1 – sin 30˚)(1+ sin30˚)
= [1 – cos230˚]/[1 – sin230˚]
= [1 – (√3/2)2]/[1 – (1/2)2]
= [1 – 3/4]/[1 – 1/4]
= [4 – 3/4]/[4-1/4]
=[ 1/4]/[3/4]
= 1/3
= RHS
- If B = 15˚, prove that 4 sin 2B.cos 4B.sin 6B = 1
Solution:
If B = 15˚, we have to prove that 4sin 2B.cos 4B.sin 6B = 1
LHS = 4sin 2B.cos 4B.sin 6B
= 4sin 2(15˚).cos 4(15˚).sin 6(15˚)
= 4 sin(30˚).cos(60˚).sin(90˚)
= 4(1/2).(1/2).(1)
= 4(1/4)
= 1
= RHS
- If A = 60˚ and B = 30˚ then prove that tan(A – B) = tanA – tanB/1 + tanAtanB
Solution:
If A = 60˚ and B = 30˚ then we have to prove that tan(A – B) = tanA – tanB/1 + tanAtanB
LHS = tan(A – B)
= tan(60˚ – 30˚)
= tan30˚
= 1/√3
RHS = tanA – tanB/1 + tanAtanB
= tan60˚ – tan30˚/1 + tan60˚tan30˚
= [√3 – 1/√3]/[1 + √3.(1/√3)]
= (2√3/3)/(1+1)
= √3/3
= 1/√3
Therefore, LHS = RHS