Trigonometry revision 4 QDHS G11 MAA HL 1st Term Chinese name: English name: Homeroom: #8 Math Class: 1. A triangle has sides of length n2 + n + 1 , (2n + 1) and n2 − 1 where n > 1. (a) Explain why the side n2 + n + 1 must be the longest side of the triangle. (b) Show that the largest angle, θ, of the triangle is 120◦ . .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 2. In the right circular cone below, O is the centre of the base which has radius 6 cm. The points B and C are on the circumference of the base of the cone. The height AO of the cone is 8 cm and the angle BÔC is 60◦ . Calculate the size of the angle ∠BAC. Page 1 of 8 Trigonometry revision 4 QDHS G11 MAA HL 1st Term #8 .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 3. Consider the function f : x → π 4 − arccos x. (a) Find the largest possible domain of f . (b) Determine an expression for the inverse function, f −1 , and write down its domain. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 4. Show that cos A + sin A = sec 2A + tan 2A. cos A − sin A .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. Page 2 of 8 QDHS G11 MAA HL 1st Term 5. (a) (i) Express cos π (ii) Hence solve 6 + √ Trigonometry revision 4 #8 x in the form a cos x − b sin x where a, b ∈ R. 3 cos x − sin x = 1 for 0 x 2π. (b) Let p(x) = 2x3 − x2 − 2x + 1. (i) Show that x = 1 is a zero of p. (ii) Hence find all the solutions of 2x3 − x2 − 2x + 1 = 0. (iii) Express sin 2θ cos θ + sin2 θ in terms of sin θ. (iv) Hence solve sin 2θ cos θ + sin2 θ = 1 for 0 θ 2π. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 6. (a) Prove the trigonometric identity sin(x + y) sin(x − y) = sin2 x − sin2 y. (b) Given f (x) = sin x + π6 sin x − π6 , x ∈ [0, π], find the range of f . (c) Given g(x) = csc x + π6 csc x − π6 , x ∈ [0, π], x ∕= π6 , x ∕= 5π 6 , find the range of g. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. Page 3 of 8 QDHS G11 MAA HL 1st Term Trigonometry revision 4 #8 7. The first three terms of a geometric sequence are sin x, sin 2x and 4 sin x cos2 x, − π2 < x < π2 . (a) Find the common ratio r. (b) Find the set of values of x for which the geometric series sin x + sin 2x + 4 sin x cos2 x + . . . converges. Consider x = arccos 1 4 , x > 0. (c) Show that the sum to infinity of this series is √ 15 2 . .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 8. Compactness is a measure of how compact an enclosed region is. 4A The compactness, C, of an enclosed region can be defined by C = πd 2 , where A is the area of the region and d is the maximum distance between any two points in the region. For a circular region, C = 1. Consider a regular polygon of n sides constructed such that its vertices lie on the circumference of a circle of diameter x units. n 2π (a) If n > 2 and even, show that C = sin . 2π n If n > 1 and odd, it can be shown that C = n sin 2π n π . π (1+cos n ) (b) Find the regular polygon with the least number of sides for which the compactness is more than 0.99. (c) Comment briefly on whether C is a good measure of compactness. Page 4 of 8 QDHS G11 MAA HL 1st Term Trigonometry revision 4 #8 .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 9. In triangle ABC, 3 sin B + 4 cos C = 6, 4 sin C + 3 cos B = 1. (a) Show that sin(B + C) = 12 . Robert conjectures that ∠CAB can have two possible values. (b) Show that Robert’s conjecture is incorrect by proving that CAB has only one possible value. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. Page 5 of 8 QDHS G11 MAA HL 1st Term Trigonometry revision 4 #8 10. In a triangle ABC, AB = 4 cm, BC = 3 cm and ∠BAC = π9 . (a) Use the cosine rule to find the two possible values for AC. (b) Find the difference between the areas of the two possible triangles ABC. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. Questions continued on next page! Page 6 of 8 Trigonometry revision 4 QDHS G11 MAA HL 1st Term 11. (Challeng ∗∗) (a) Show that cos 15◦ = √ 3+1 √ 2 2 #8 and find a similar expression for sin 15◦ . (b) Show that cos α is a root of the equation 4x3 − 3x − cos 3α = 0, and find the other two roots in terms of cos α and sin α. √ (c) Use parts (i) and (ii) to solve the equation y 3 − 3y − 2 = 0, giving your answers in surd form. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. Question continued on next page! Page 7 of 8 QDHS G11 MAA HL 1st Term Trigonometry revision 4 #8 12. (Challeng ∗∗) √ √ (a) Find integers m and n such that 3 + 2 2 = m + n 2. (b) Let f(x) = x4 − 10x2 + 12x − 2. Given that the equation f(x) = 0 has four real roots, explain why f(x) can be written in the form f(x) = x2 + sx + p x2 − sx + q for some real constants s, p and q, and find three equations for s, p and q. Show that 2 s2 s2 − 10 + 8s2 − 144 = 0 and find the three possible values of s2 . Use the smallest of these values of s2 to solve completely the equation f(x) = 0, simplifying your answers as far as you can. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. .............................................................................................. 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