Chapter 8: Introduction to Trigonometry

Key Formulas:
• sin θ = Perpendicular/Hypotenuse, cos θ = Base/Hypotenuse, tan θ = P/B
• cosec θ = 1/sinθ, sec θ = 1/cosθ, cot θ = 1/tanθ
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

Exercise 8.1

Q1. In ΔABC, right-angled at B, AB=24cm, BC=7cm. Find sin A, cos A, sin C, cos C.

Solution

AC = √(24²+7²) = √(576+49) = √625 = 25cm
sin A = BC/AC = 7/25
cos A = AB/AC = 24/25
sin C = AB/AC = 24/25
cos C = BC/AC = 7/25

Q3. If sin A = 3/4, find cos A and tan A.

Solution

sin²A+cos²A=1
9/16+cos²A=1
cos²A=7/16
cos A=√7/4
tan A=sin A/cos A = 3/√7 = 3√7/7

Q4. Given 15 cot A = 8, find sin A and sec A.

Solution

cot A = 8/15 = Adjacent/Opposite
Hypotenuse = √(8²+15²) = √(64+225) = √289 = 17
sin A = 15/17
sec A = 17/8

Q5. Given sec θ = 13/12, find all trigonometric ratios.

Solution

sec θ=13/12 → cos θ=12/13
sin²θ=1-144/169=25/169 → sin θ=5/13
tan θ=5/12
cosec θ=13/5
cot θ=12/5

Q9. ΔPQR, right-angled at Q, PR+QR=25cm, PQ=5cm. Find sin P, cos P, tan P.

Solution

PQ=5, let QR=x, PR=25-x
5²+x²=(25-x)²
25+x²=625-50x+x²
50x=600 → x=12
QR=12, PR=13
sin P=12/13, cos P=5/13, tan P=12/5

Exercise 8.2

Q1. Evaluate: sin 60° cos 30° + sin 30° cos 60°

Solution

= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4 = 1

Q2. Evaluate: 2 tan² 45° + cos² 30° - sin² 60°

Solution

= 2(1) + 3/4 - 3/4 = 2

Q3. Evaluate: (cos 0°+sin 45°)/(sin 0°+cos 45°)

Solution

= (1+1/√2)/(0+1/√2)
= (√2+1)/√2 × √2/√2 = √2+1

Q4. Evaluate: (5cos² 60°+4sec² 30°-tan² 45°)/(sin² 30°+cos² 30°)

Solution

= (5/4+16/3-1)/1
= (15+64-12)/12
= 67/12

Q6. Evaluate: (sin 30°/tan 45°)+(cos 60°/tan 30°)

Solution

= (1/2)/1 + (1/2)/(1/√3)
= 1/2 + √3/2
= (1+√3)/2

Exercise 8.3

Q1. Evaluate: (i) sin² 30°+cos² 30° (ii) sin² 60²+cos² 60°

Solution

(i) (1/2)²+(√3/2)² = 1/4+3/4 = 1
(ii) (√3/2)²+(1/2)² = 3/4+1/4 = 1

Q4. Evaluate: cos² 0°+cos² 30°+cos² 45°+cos² 60°+cos² 90°

Solution

= 1+3/4+1/2+1/4+0
= 1+0.75+0.5+0.25
= 5/2

Exercise 8.4

Q1. Express sin 67°+cos 75° in terms of angles between 0° and 45°.

Solution

sin 67°=sin(90°-23°)=cos 23°
cos 75°=cos(90°-15°)=sin 15°
∴ sin 67°+cos 75° = cos 23°+sin 15°

Q3. If tan 2A=cot(A-18°), find A.

Solution

tan 2A=cot(A-18°)=tan(90°-(A-18°))
tan 2A=tan(108°-A)
2A=108°-A
3A=108°
A=36°

Q4. If tan A=cot B, prove A+B=90°.

Solution

tan A=cot B=tan(90°-B)
A=90°-B
∴ A+B=90°

Q5. If sec 4A=cosec(A-20°), find A.

Solution

sec 4A=cosec(A-20°)
cos(4A)=sin(A-20°)
sin(90°-4A)=sin(A-20°)
90°-4A=A-20°
110°=5A
A=22°

Q6. In ΔABC, show sin((B+C)/2)=cos(A/2).

Solution

A+B+C=180°
B+C=180°-A
(B+C)/2=90°-A/2
sin((B+C)/2)=sin(90°-A/2)=cos(A/2)

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