Show that, after making the substitution x = 2 sin θ 3 , the equation x 3 - x + 1 6 3 = 0 can be written in the form sin 3 θ = 3 4 .

Given: x 3 - x + 1 6 3 = 0 Let x = 2 sin θ 3 ⇒ 2 sin θ 3 3 - 2 sin θ 3 + 1 6 3 = 0 ⇒ 8 sin 3 θ 3 3 - 2 sin θ 3 + 1 6 3 = 0 ⇒ 16 sin 3 θ - 12 sin θ + 3 = 0 ⇒ - 4 3 sin θ - 4 sin 3 θ + 3 = 0 ⇒ - 4 sin 3 θ + 3 = 0 ⇒ sin 3 θ = 3 4

Show that, after making the substitution x = 2 sin θ 3 , the equation x 3 - x + 1 6 3 = 0 can be written in the form sin 3 θ = 3 4 . Hence solve the equation x 3 - x + 1 6 3 = 0 , giving your answer correct to 3 significant figures.

Given: x 3 - x + 1 6 3 = 0 Let x = 2 sin θ 3 ⇒ 2 sin θ 3 3 - 2 sin θ 3 + 1 6 3 = 0 ⇒ 8 sin 3 θ 3 3 - 2 sin θ 3 + 1 6 3 = 0 ⇒ 16 sin 3 θ - 12 sin θ + 3 = 0 ⇒ - 4 3 sin θ - 4 sin 3 θ + 3 = 0 ⇒ - 4 sin 3 θ + 3 = 0 ⇒ sin 3 θ = 3 4 Now, sin 3 θ = 3 4 3 θ = 48 . 59 ° ⇒ θ = 16 . 2 ° ⇒ 2 sin θ 3 = 0 . 322 or ⇒ 3 θ = 180 ° - 48 . 59 ° = 131 . 40 ° ⇒ θ = 43 . 8 ° ⇒ 2 sin θ 3 = 0 . 799 or 3 θ = 131 . 40 ° - 360 ° ⇒ θ = - 76 . 2 ° ⇒ 2 sin θ 3 = - 1 . 12

Prove the identity 1 sin x + 30 ° + cos x + 60 ° ≡ sec x .

To prove: 1 sin x + 30 ° + cos x + 60 ° ≡ sec x . LHS : 1 sin x + 30 ° + cos x + 60 ° ≡ 1 sin x cos 30 ° + cos x sin 30 ° + cos x cos 60 ° - sin x sin 60 ° ≡ 1 3 2 sin x + 1 2 cos x + 1 2 cos x - 3 2 sin x ≡ 1 cos x ≡ sec x Hence, proved.

Prove the identity 1 sin x + 30 ° + cos x + 60 ° ≡ sec x . Hence solve the equation 1 sin x + 30 ° + cos x + 60 ° = 7 - tan 2 x for 0 ° < x < 360 ° .

To prove: 1 sin x + 30 ° + cos x + 60 ° ≡ sec x . LHS : 1 sin x + 30 ° + cos x + 60 ° ≡ 1 sin x cos 30 ° + cos x sin 30 ° + cos x cos 60 ° - sin x sin 60 ° ≡ 1 3 2 sin x + 1 2 cos x + 1 2 cos x - 3 2 sin x ≡ 1 cos x ≡ sec x Now, sec x = 7 - tan 2 x ⇒ sec x = 7 - sec 2 x - 1 ⇒ sec 2 x + sec x - 8 = 0 Let u = sec x ⇒ u 2 + u - 8 = 0 ⇒ u = - 1 ± 1 2 - 4 × 1 × ( - 8 ) 2 × 1 = - 1 ± 33 2 ⇒ sec x = - 1 ± 33 2 ⇒ cos x = 2 - 1 + 33 or cosx = 2 - 1 - 33 ⇒ x = 60 , x = 104 . 5 , x = 255 . 5 , x = 300

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3

Attempt the free practice questions on Chapter 3: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3 with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3 with Hints & Solutions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 12

Express $4 \cos θ + 6 \sin θ$ in the form $R \cos (θ - α),$ where $R > 0$ and $0 < α < \frac{π}{2} .$ Give the exact value of $R$ and the value of $α$ correct to $2$ decimal places. Write down the greatest value of ${(\frac{2 \sin 2 θ - 3 \cos 2 θ + 3}{\sin θ})}^{2} .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 13

Express $4 \sin θ - 6 \cos θ$ in the form $R \sin (θ - α),$ where $R > 0$ and $0 ° < α < 90 °$ Give the exact value of $R$ and the value of $α$ correct to $2$ decimal places.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 13

Express $4 \sin θ - 6 \cos θ$ in the form $R \sin (θ - α),$ where $R > 0$ and $0 ° < α < 90 °$ Give the exact value of $R$ and the value of $α$ correct to $2$ decimal places. Solve the equation $4 \sin θ - 6 \cos θ = 3$ for $0 ° ⩽ θ ⩽ 360 ° .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 13

Express $4 \sin θ - 6 \cos θ$ in the form $R \sin (θ - α),$ where $R > 0$ and $0 ° < α < 90 °$ Give the exact value of $R$ and the value of $α$ correct to $2$ decimal places

.Find the greatest and least possible values of $(4 \sin θ - 6 \cos θ)^{2} + 8$ as $θ$ varies.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 14

Show that, after making the substitution $x = \frac{2 \sin θ}{\sqrt{3}}$ , the equation $x^{3} - x + \frac{1}{6} \sqrt{3} = 0$ can be written in the form $\sin 3 θ = \frac{3}{4}$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 14

Show that, after making the substitution $x = \frac{2 \sin θ}{\sqrt{3}}$ , the equation $x^{3} - x + \frac{1}{6} \sqrt{3} = 0$ can be written in the form $\sin 3 θ = \frac{3}{4}$ .

Hence solve the equation $x^{3} - x + \frac{1}{6} \sqrt{3} = 0$ , giving your answer correct to $3$ significant figures.

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 15

Prove the identity $\frac{1}{\sin (x + 30 °) + \cos (x + 60 °)} \equiv \sec x .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 3 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 3>Q 15

Prove the identity $\frac{1}{\sin (x + 30 °) + \cos (x + 60 °)} \equiv \sec x .$ Hence solve the equation $\frac{1}{\sin (x + 30 °) + \cos (x + 60 °)} = 7 - \tan^{2} x$ for $0 ° < x < 360 ° .$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Trigonometry, Exercise 8: END-OF-CHAPTER REVIEW EXERCISE 3 with Hints & Solutions

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