Page 58 - mathsvol1ch1to3ans

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Solution:

2
2
(ii) sin B + sin C = 1
√ B C
(iii) cos B − cos C = −1 + 2 2 cos sin .
2 2
Exercise 3.8:
1. Find the principal solution and general solutions of the following:
1 √ 1
(i) sin θ = −√ (ii) cot θ = 3 (iii) tan θ = −√ .
2 3
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Solution:
1
(i) sin θ = −√ < 0 principal value lies in the IV quadrant
2
1 π π
sin θ = −√ = − sin θ = nπ + (−1) n − . n ∈ Z
2 4 4
√ 1
(ii) cot θ = 3 ⇒ tan θ = √ > 0 principal value lies in the I quadrant
3
1 π π
tan θ = √ = tan θ = nπ + n ∈ Z
3 6 6
1
(iii) tan θ = −√ < 0 principal value lies in the IV quadrant
3
1 π π
tan θ = −√ = − tan θ = nπ − n ∈ Z
3 6 6
◦
2. Solve the following equations for which solutions lies in the interval 0 ≤ θ < 360 ◦
4
2

2
2
2
4
(i) sin x = sin x sin x − sin x = 0 ⇒ sin x sin x − 1 = 0
sin (x) = 1, sin (x) = −1, sin (x) = 0
π 3π
sin (x) = 1 ⇒ x = : sin (x) = −1 ⇒ x = : sin (x) = 0 ⇒ x = 0 x = π
2 2 1
2
2
(ii) 2 cos x + 1 = −3 cos x 2 cos x + 3 cos x + 1 = 0 ⇒ cos (x) = − , cos (x) = −1
2
1 2π 4π
cos (x) = − ⇒ x = , : cos (x) = −1 ⇒ x = π
2 3 3
1
2
2
(iii) 2 sin x + 1 = 3 sin x 2 sin x − 3 sin x + 1 = 0 ⇒ sin (x) = 1, sin (x) =
2
π 1 π 5π
sin (x) = 1 ⇒ x = : sin (x) = ⇒ x = :
2 2 6 6
(iv) cos 2x = 1 − 3 sin x.

3
2
1 − 2 sin x = 1 − 3 sin x ⇒ sin x sin x − ⇒ sin x = 0 ⇒ x = 0, π
2
3. Solve the following equations:
(i) sin 5x − sin x = cos 3x
2 sin 2x cos 3x = cos 3x ⇒ cos 3x(2 sin 2x − 1) = 0
π π
When cos 3x = 0, 3x = (2n + 1) and hence x = (2n + 1)
2 6