Soal Trigonometri - Copy Backup

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EXERCISE 8B 1 Use the compound-angle formulae to write the following as surds.

(i) sin75° = sin(45° + 30°) (ii) cos 135° = cos(90° + 45°)

(iii) tan15° = tan(45° – 30°) (iv) tan75° = tan(45° + 30°)

2 Expand each of the following expressions.

(i) sin(θ + 45°) (ii) cos(θ – 30°)

(iii) sin(60° – θ) (iv) cos(2θ + 45°)

(v) tan(θ + 45°) (vi) tan(θ – 45°)

3 Simplify each of the following expressions.

(i) sin2θcosθ – cos2θ sinθ

(ii) coszcos3z – sinz sin3z

(iii) sin120°cos60° + cos120°sin60°

(iv) cosθcosθ – sinθsin θ

4 Solve the following equations for values of θ in the range 0° θ 180°.

(i) cos(60° + θ) = sin θ

(ii) sin(45° – θ) = cosθ

(iii) tan(45° + θ) = tan(45° – θ)

(iv) 2sinθ = 3cos(θ – 60°)

(v) sinθ = 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 multiples of π where appropriate.)

(i) sin(θ + ) = cosθ

(ii) 2cos(θ – ) = cos(θ + )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. The angles α and β and the point P are shown in the diagram.

π–

2

π–

3

π

–4

y

xO

α β

M(0, 2)

N(3, –2)

L(–2, 1)

P



7 (i) Find ∫ x coskx dx , where k is a non-zero constant.

(ii) Show that

cos(A – B ) – cos(A + B ) = 2sinA sinB .

Hence express 2sin5x sin3x as the difference of two cosines.

(iii) Use the results in parts (i) and (ii) to show that

∫ 0 x sin5x sin3x dx = .

[MEI]

8 (i) Use the formulae for cos(θ + z) and cos(θ – z) to prove that

cos(θ – z) – cos(θ + z) = 2sinθ sinz. ∗

Prove also that sin (π – θ) = sin θ.

In triangle PQR, angle P = 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.

(ii) Show that PQ = 2 sin(α + π).

By finding RS in terms of α, deduce that the area A of the triangle is given by

A = sin(α + π) sinα.

Find the value of α for which the area A is a maximum. [You may find the

result ∗ helpful.]

(iii) Expand sin(α + 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, sinθ 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]

1–

6

1–

6

1–

6

1–

6

π – 2–––––

16

π–4

R

SP Q

απ1

6

(i) Show that sinα = and write down the value of cosα.

(ii) Find the values of sin β and cos β .

(iii) Show that sinLMN = .

(iv)

Show that tan

LNM = .

11

––27

11–––

5 5

2––

5



EXERCISE 8C 1 Solve the following equations for 0° θ 360°.

(i) 2sin2θ = cosθ (ii) tan2θ = 4tanθ (iii) cos2θ + sinθ = 0

(iv) tanθ tan2θ = 1 (v) 2cos2θ = 1 + cos θ

2 Solve the following equations for –π θ π.

(i) sin2θ = 2sinθ (ii) tan2θ = 2tanθ (iii) cos2θ – cosθ = 0

(iv) 1 + cos2θ = 2sin2 θ (v) sin4θ = cos2θ

(Hint: Write the expression in part (v) as an equation in 2θ.)

3 By first writing sin3θ as sin(2θ + θ), express sin3θ in terms of sin θ.

Hence solve the equation sin3θ = sinθ for 0 θ 2π.

4 Solve cos3θ = 1 – 3cosθ for 0° θ 360°.

1 + cos2θ5 Simplify –––––––– .

sin2θ

6 Express tan3θ in terms of tanθ.

1 – tan2 θ7 Show that –––––––– = cos 2θ.

1 + tan2 θ

8 (i) Show that tan( + θ)tan( – θ) = 1.

(ii) Given that tan 26.6° = 0.5, solve tanθ = 2 without using your calculator.

Give θ to 1 decimal place, where 0° θ 90°.

9 (i) Sketch on the same axes the graphs of

y = cos2x and y = 3sinx – 1 for 0 x 2π.

(ii) Show that these curves meet at points whose x co-ordinates are solutionsof the equation 2sin2 x + 3sinx – 2 = 0.

(iii) Solve this equation to find the values of x in terms of π for 0 x 2π.

[MEI]

π–

4

π–

4



EXAMPLE 8.6 Solve sin3θ + sinθ = 0 for 0° θ 360°.

SOLUTION

Using

α + β α – β sinα + sin β = 2sin

(––––

)cos

(––––

)2 2

and putting α = 3θ and β = θ gives

sin3θ + sinθ = 2 sin 2θcos θ

so the equation becomes

2sin2θcosθ = 0

⇒ cosθ = 0 or sin2θ = 0.

From the graphs for y = cosθ and y = sinθ

cosθ = 0 gives θ = 90° or 270°

sin2θ = 0 gives 2θ = 0°, 180°, 360°, 540° or 720°

so 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 – sin2θ

(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 + cos2θ = cos θ.

sin5θ + sin3θ3 Simplify ––––––––––––.sin5θ – sin3θ

4 Solve the equation sin3θ – sinθ = 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 + sinθ = 4sinθcos2 θ.

(ii) Hence show that sin3θ = 3sinθ – 4sin3 θ.

You should only list

each root once in the

final answer.

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