Page 40 - mathsvol1ch1to3ans

40

1
(i) cos θ = − , θ lies in the III quadrant.
2

q √ √
2
Solution:sin θ = ± 1 − − 1 2 = − 3 tan θ = 3; cosecθ = − √ ; sec θ = −2; cot θ = √ 1
2 2 3 3
2
(ii) cos θ = , θ lies in the I quadrant.
3
√
r
4 5
Solution:sin θ = ± 1 − =
9 3
q 2 √
(i) cos θ = − 1 θ lies in the III quadrant. sin θ = ± 1 − − 1 = − 3
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2 2 2
√
3 2 √ 1
(3)(i) sin θ = − ; cosec θ = −√ ; sec θ = −2; tan θ = 3; cot θ = √
2
√ 3 √ 3
5 3 3 5 2
(ii) sin θ = ; cosec θ = −√ ; sec θ = ; tan θ = ; cot θ = √
3
√ 5 2 2 √ 5
5 3 3 2 5
(iii) cos θ = ; cosec θ = − ; sec θ = √ ; tan θ = −√ ; cot θ = −
3 2 5 5 √ 2
√ 1 1 2 5
(iv) sec θ = − 5; cos θ = −√ ; cot θ = − ; sin θ = √ ; cosec θ = ;
5 2 5 2
5 12 13 12 5
(v) cos θ = ; sin θ = − ; cosec θ = − ; tan θ = − ; cot θ = − ;
13 13 12 5 12
◦
◦
cot(180 + θ) sin(90 − θ) cos(−θ)
2
4. Prove that = cos θ tan θ.
◦
◦
sin(270 + θ) tan(−θ)cosec(360 + θ)
◦
◦
cot(180 + θ) sin(90 − θ) cos(−θ) cot θ cos θ cos θ
=
◦
◦
sin(270 + θ) tan(−θ)cosec(360 + θ) − cos θ − tan θcosecθ
3
cos θ cos θ sin θ
=
sin θ sin θ
2
= cos θ cot θ
3
2
◦
◦
5. Find all the angles between 0 and 360 which satisfy the equation sin θ = .
√ 4
3 π 2π 4π 5π
Solution: sin θ = ± . Hence the angles are , , ,
2 3 3 3 3
6. Show that sin 2 π + sin 2 π + sin 2 7π + sin 2 4π = 2.
9
9
18
Solution: sin 4π = cos π − 4π 18
9 2 9
= sin 2 π + sin 2 π + sin 2 7π + cos 2 π − 4π
18 9 18 2 9
= sin 2 π + sin 2 π + sin 2 7π + cos 2 π
18 9 18 18
= 1 + sin 2 π + sin 2 7π
9 18
= 1 + sin 2 π + cos 2 π − 7π
9 2 18
= 1 + sin 2 π + cos 2 π
9 9
= 1 + 1
= 2
Exercise 3.4: