EXERCISE 8B
1
Use the compound-angle formulae to write the following as surds. (i) sin75° = sin( (ii) cos sin(45 45°° + 30°) 30°) cos 135° 135° = cos(9 cos(90° 0° + 45°) 45°) tan 15° 15° = tan( tan(45 45°° – 30°) 30°) tan 75° 75° = tan( tan(45 45°° + 30°) 30°) (iii) tan (iv) tan
2
Expand each of the following expressions. (i) sin(θ + 45°) (ii) cos(θ – 30°) sin(60° 0° – θ) (iii) sin(6 (iv) cos(2θ + 45°) (v) tan(θ + 45°) (vi) tan(θ – 45°)
3
Simplify each of the following expressions. (i) sin2θ cos θ – cos2θ sin θ (ii) cos z cos3z – sin z sin3z (iii) sin sin 120°cos60° 120°cos60° + cos120 cos120°° sin60° (iv) cos θ cos θ – sin θ sin θ
4
Solve the following equations for values of θ in the range 0° θ 180°. cos(60 60°° + θ) = sin θ (i) cos( (ii) sin( sin(45 45°° – θ) = cos θ tan(45 45°° + θ) = tan(4 tan(45° 5° – θ) (iii) tan( cos(θ – 60°) (iv) 2sin θ = 3 cos( (v) sin θ = cos(θ + 120°)
5
Solve the following equations for values of θ in the range 0 θ π. (When the range is given in radians, the solutions should be in radians, using using multipl multiples es of π where appropriate.) π (i) sin θ + – = cos θ 4 π π (ii) 2cos θ – – = cos θ + – 3 2
(
)
(
6
)
(
)
Calculators are not to be used in this question. The diagram shows three points L(–2, 1), M(0, 2) and N(3, –2) joined to form a triangle. triangle. The The angles angles α and β and the point P are shown in the diagram. y
M(0, 2) L(–2, 1)
α
β P
O
x
N(3, –2)
(i)
2 Show that sin α = –– and write down the value of cos α. 5
(ii)
Find the values of sin β and cos β .
(iii)
(iv) 7 (i) (ii)
11 Show that sin LMN = ––– . 5 5 11 Show that tan LNM = –– . 27 Find x coskx dx , where k is a non-zero constant. Show that
∫
cos(A – B ) – cos(A + B ) = 2sin A sin B . (iii)
Hence express 2sin5x sin3x as the difference of two cosines. Use the results in parts (i) and (ii) to show that π – 4
∫
–2 x sin5x sin3x dx = π ––––– . 16 0
8 (i)
[MEI]
Use the formulae for cos(θ + z) and cos(θ – z) to prove that cos(θ – z) – cos(θ + z) = 2sin θ sin z.
∗
Prove also that sin (π – θ) = sin θ. 1
In triangle PQR, angle P = –6 π radians, angle Q = α radians, and QR = 1 unit. The point S is at the foot of the perpendicular from R to PQ, as shown in the diagram. R
P
1 6
α
π
Q S
(ii)
Show that PQ = 2 sin(α + –6 π). By finding RS in terms of α, deduce that the area A of the triangle is given by 1
1 A = sin(α + –6 π) sin α.
Find the value of α for which the area A is a maximum. [You may find the result ∗ helpful.] (iii)
1 Expand sin(α + –6 π), and hence show that, for small values of α,
A ≈ p α + q α2, where p and q are contants to be determined.
[For small θ, sin θ
θ and cos θ
1.]
Find the value of this expression when α = 0.1, and find also the corresponding value of A given by the expression in part (ii).
[MEI]
EXERCISE 8C
1
2
Solve the following equations for 0° θ 360°. (i) 2sin2θ = cos θ (ii) tan2θ = 4 tan θ (iv) tan θ tan2θ = 1 (v) 2cos2θ = 1 + cos θ
(iii)
cos2θ + sin θ = 0
Solve the following equations for –π θ π. (i) sin2θ = 2sin θ (ii) tan2θ = 2 tan θ 2 (iv) 1 + cos 2θ = 2sin θ (v) sin4θ = cos2θ
(iii)
cos2θ – cos θ = 0
(Hint: Write the expression in part (v) as an equation in 2θ.) 3
By first writing sin 3θ as sin(2θ + θ), express sin3θ in terms of sin θ. Hence solve the equation sin3θ = sin θ for 0 θ 2π.
4
Solve cos3θ = 1 – 3cos θ for 0° θ 360°.
5
+ cos 2θ . Simplify 1–––––––– sin2θ
6
Express tan3θ in terms of tan θ.
8
– tan2 θ = cos 2θ. Show that 1–––––––– 1 + tan2 θ π π (i) Show that tan – + θ tan – – θ = 1. 4 4 (ii) Given that tan 26.6° = 0.5, solve tan θ = 2 without using your calculator. Give θ to 1 decimal place, where 0° θ 90°.
9
(i)
7
(
) (
)
Sketch on the same axes the graphs of y = cos 2x and y = 3 sin x – 1 for 0 x 2π.
(ii)
Show that these curves meet at points whose x co-ordinates are solutions of the equation 2 sin2 x + 3sin x – 2 = 0.
(iii)
Solve this equation to find the values of x in terms of π for 0 x 2π. [MEI]
EXAMPLE 8.6
Solve sin3θ + sin θ = 0 for 0° θ 360°. SOLUTION
Using
(
) (
)
α + β α – β sin α + sin β = 2 sin –––– cos –––– 2 2
and putting α = 3θ and β = θ gives sin3θ + sin θ = 2 sin 2θ cos θ so the equation becomes 2sin2θ cos θ = 0 cos θ = 0
⇒
or
sin2 θ = 0.
From the graphs for y = cos θ and y = sin θ cos θ = 0 gives θ = 90° or 270° sin2θ = 0 gives 2θ = 0°, 180°, 360°, 540° or 720° so
You should only list each root once in the final answer.
θ = 0°, 90°, 180°, 270° or 360°.
The complete set of roots in the range given is θ = 0°, 90°, 180°, 270°, 360°. EXERCISE 8D
The questions in this exercise relate to enrichment material.
1
Factorise the following expressions. (i) sin4θ – sin 2θ (ii) cos5θ + cos θ (iii) cos7θ – cos3θ (iv) cos(θ + 60°) + cos( θ – 60°) (v) sin(3θ + 45°)+ sin(3θ – 45°)
2
Factorise cos4θ + cos 2θ. Hence, for 0° θ 180°, solve cos4θ + cos 2θ = cos θ .
3
sin5θ + sin 3θ. Simplify –––––––––––– sin5θ – sin 3θ
4
Solve the equation sin3θ – sin θ = 0 for 0 θ 2π.
5
Factorise sin(θ + 73°) – sin(θ + 13°) and use your result to sketch the graph of y = sin(θ + 73°) – sin( θ + 13°).
6
Prove that sin2 A – sin2 B = sin(A – B ) sin(A + B ).
7
(i )
Use a suitable factor formula to show that sin3θ + sin θ = 4sin θ cos 2 θ .
(ii)
Hence show that sin3θ = 3 sin θ – 4sin3 θ .
