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1Exercise 3.1
Convert the following to radians:<br>(i) 25° (ii) 240° (iii) -47°30'.
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Use 1° = π/180 radians.
(i) 25° = 25 × π/180 = 5π/36 rad
(ii) 240° = 240 × π/180 = 4π/3 rad
(iii) -47°30' = -(47 + 30/60)° = -47.5° = -47.5 × π/180 = -95π/360 = -19π/72 rad
2Exercise 3.1
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
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360 revolutions in 60 seconds ⇒ 6 revolutions per second.
One revolution = 2π radians.
Radians per second = 6 × 2π = 12π rad/s
3Exercise 3.1
Find the value of sin 75° + cos 75° and tan 15°.
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sin 75° = sin(45°+30°) = sin45 cos30 + cos45 sin30
= (1/√2)(√3/2) + (1/√2)(1/2) = (√3+1)/(2√2)
cos 75° = cos(45°+30°) = cos45 cos30 - sin45 sin30
= (1/√2)(√3/2) - (1/√2)(1/2) = (√3-1)/(2√2)
sin 75° + cos 75° = (√3+1+√3-1)/(2√2) = 2√3/(2√2) = √3/√2 = √6/2
tan 15° = tan(45°-30°) = (1 - 1/√3)/(1 + 1/√3) = (√3-1)/(√3+1) = (√3-1)²/(3-1)
= (3 - 2√3 + 1)/2 = 2 - √3
4Exercise 3.1
Prove that cos 2x = (1 - tan²x)/(1 + tan²x).
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RHS = (1 - tan²x)/(1 + tan²x)
= (1 - sin²x/cos²x)/(1 + sin²x/cos²x)
= ((cos²x - sin²x)/cos²x)/((cos²x + sin²x)/cos²x)
= (cos²x - sin²x)/(cos²x + sin²x)
= (cos²x - sin²x)/1
= cos 2x = LHS. Hence proved.
5Exercise 3.1
If tan A = 3/4 and tan B = 5/12, find tan(A + B).
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tan(A+B) = (tan A + tan B)/(1 - tan A tan B)
= (3/4 + 5/12)/(1 - (3/4)(5/12))
= (9/12 + 5/12)/(1 - 15/48)
= (14/12)/(33/48)
= (7/6)/(11/16)
= (7/6) × (16/11) = 56/33
6Exercise 3.1
Find the general solution of the equation sin x = 1/2.
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sin x = 1/2 ⇒ x = π/6 in [0, π/2].
General solution:
x = nπ + (-1)ⁿ(π/6), n ∈ Z