Show that sin θ sin θ + cos θ + cos θ sin θ - cos θ ≡ 1 sin 2 θ - cos 2 θ .

Given that, the trigonometric function is LHS = sin θ sin θ + cos θ + cos θ sin θ - cos θ ≡ 1 sin 2 θ - cos 2 θ = sin θ ( sin θ - cos θ ) ( sin θ + cos θ ) ( sin θ - cos θ ) + cos θ ( sin θ + cos θ ) ( sin θ + cos θ ) ( sin θ - cos θ ) = sin θ ( sin θ - cos θ ) ( sin θ + cos θ ) ( sin θ - cos θ ) + cos θ ( sin θ + cos θ ) ( sin θ + cos θ ) ( sin θ - cos θ ) = sin 2 θ - sin θ cos θ + sin θ cos θ + cos 2 θ sin 2 θ - cos 2 θ We know that, sin 2 x + cos 2 x ≡ 1 = 1 - sin θ cos θ + sin θ cos θ sin 2 θ - cos 2 θ = 1 sin 2 θ - cos 2 θ = RHS Hence, shown.

Hence, solve the equation sin θ sin θ + cos θ + cos θ sin θ - cos θ = 3 for 0 ° ⩽ θ ⩽ 360 ° .

Given that, the trigonometric expression is sin θ sin θ + cos θ + cos θ sin θ - cos θ = 3 for 0 ° ⩽ θ ⩽ 360 ° ⇒ sin θ sin θ + cos θ + cos θ sin θ - cos θ = 1 sin 2 θ - cos 2 θ = 3 ⇒ 3 sin 2 θ - cos 2 θ = 1 We know that, sin 2 x + cos 2 x ≡ 1 ⇒ 3 1 - cos 2 θ - cos 2 θ = 1 ⇒ 3 - 6 cos 2 θ = 1 ⇒ cos θ = ± 1 3 If cos θ = 1 3 then θ = 54 . 73 … ° As cos is positive in the first and fourth quadrants: The second solution is 360 ° - 54 . 73 … ° = 305 . 26 ° If cos θ = - 1 3 then θ = 125 . 26 … ° As cos is negative in the second and third quadrants. Using the symmetry of the sketch: The second solution is 234 . 73 … ° Thus the solutions are: 54 . 7 ° , 125 . 3 ° , 234 . 7 ° , 305 . 3 ° (to 1 decimal place).

Show that the equation 4 cos θ tan θ + 15 = 0 can be expressed as 4 sin 2 θ - 15 sin θ - 4 = 0 .

Given that, the trigonometric function is 4 cos θ tan θ + 15 = 0 We know that, tan x = sin x cos x ⇒ 4 cos 2 θ sin θ + 15 = 0 ⇒ 4 cos 2 θ + 15 sin θ = 0 We know that, sin 2 x + cos 2 x = 1 ⇒ 4 1 - sin 2 θ + 15 sin θ = 0 ⇒ 4 - 4 sin 2 θ + 15 sin θ = 0 ∴ 4 sin 2 θ - 15 sin θ - 4 = 0 Hence, shown.

Hence, solve the equation 4 cos θ tan θ + 15 = 0 for 0 ° ≤ θ ≤ 360 ° .

Given that, the trigonometric equation is 4 cos θ tan θ + 15 = 0 ⇒ 4 cos 2 θ + 15 sin θ = 0 ⇒ 4 1 - sin 2 θ + 15 sin θ = 0 ⇒ 4 cos θ tan θ + 15 = 0 = 4 sin 2 θ - 15 sin θ - 4 = 0 Factorising the left-hand side of 4 sin 2 θ - 15 sin θ - 4 = 0 gives: ( 4 sin θ + 1 ) ( sin θ - 4 ) = 0 Case (i) : 4 sin θ + 1 = 0 ⇒ sin θ = - 1 4 ⇒ θ = - 14 . 47 … ° (out of range) Using the symmetry of the sketch: The first solution is 180 ° - - 14 . 47 … ° = 194 . 47 … ° The second solution is 360 ° - 14 . 47 … ° = 345 . 53 … ° Case (ii) : sin θ - 4 = 0 sin θ = 4 No solutions since - 1 ⩽ sin θ ⩽ 1 Solutions are 194 . 5 ° or 345 . 5 ° (to 1 decimal place).

The function f : x → 5 + 3 cos 1 2 x is defined for 0 ≤ x ⩽ 2 π . Solve the equation f ( x ) = 7 , giving your answer correct to 2 decimal places.

Given that, the trigonometric function is f x = 5 + 3 cos 1 2 x Also, f x = 7 ⇒ 5 + 3 cos 1 2 x = 7 ⇒ cos 1 2 x = 2 3 ⇒ x = 96 . 37 ° or 1 2 x = 360 ° - 48 . 18 ° ⇒ x = 96 . 37 ° or x = 2 360 ° - 48 . 18 ° = 623 . 64 ° Since, cos 360 ° + θ = cos θ Since, 623 . 64 ° is not in the range 0 ° ⩽ x ⩽ 360 ° . The solution of the trigonometric function is 96 . 37 ° .

The function f : x → 5 + 3 cos 1 2 x is defined for 0 ≤ x ⩽ 2 π . Sketch the graph of y = f ( x ) .

The graph of f ( x ) = cos x (period 2 π , amplitude 1 ) is transformed to the graph of f ( x ) = 5 + 3 cos 1 2 x by the following transformations: A horizontal stretch factor 2 (all x - coordinates are multiplied by 2 , the period is now 4 π ) A vertical stretch factor 3 (all the y - coordinates are multiplied by 3 , the range is now - 3 ⩽ f ( x ) ⩽ 3 ) A translation 0 5 (all the y - coordinates increases by 5 , the range is now 2 ⩽ f ( x ) ⩽ 8 ) The domain is unchanged. See sketch:

The function f : x → 5 + 3 cos 1 2 x is defined for 0 ≤ x ≤ 2 π . Explain why f has an inverse.

Given that, the function is f x = 5 + 3 cos 1 2 x for 0 ≤ x ≤ 2 π The function f x = 5 + 3 cos 1 2 x is many one function for x ∈ R . Since, the domain is restricted to 0 ≤ x ≤ 2 π , the function is one one and onto function. So, the function is bijective function and the inverse exists with this domain.

The function f : x → 5 + 3 cos 1 2 x is defined for 0 ≤ x ⩽ 2 π . Obtain an expression for f - 1 ( x ) .

Given that, the trigonometric function is f ( x ) = 5 + 3 cos 1 2 x Here, y = 5 + 3 cos 1 2 x Let x = 5 + 3 cos 1 2 y ⇒ cos 1 2 y = x - 5 3 ⇒ 1 2 y = cos - 1 x - 5 3 ⇒ y = 2 cos - 1 x - 5 3 ⇒ f - 1 ( x ) = 2 cos - 1 x - 5 3 Thus the inverse of the given function is 2 cos - 1 x - 5 3 .

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5

Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 12: END-OF-CHAPTER REVIEW EXERCISE 5

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 12: END-OF-CHAPTER REVIEW EXERCISE 5

Attempt the practice questions on Chapter 5: Trigonometry, Exercise 12: END-OF-CHAPTER REVIEW EXERCISE 5 with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 12: END-OF-CHAPTER REVIEW EXERCISE 5 with Hints & Solutions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5>Q 15

Show that $\frac{\sin θ}{\sin θ + \cos θ} + \frac{\cos θ}{\sin θ - \cos θ} \equiv \frac{1}{\sin^{2} θ - \cos^{2} θ} .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5>Q 15

Hence, solve the equation $\frac{\sin θ}{\sin θ + \cos θ} + \frac{\cos θ}{\sin θ - \cos θ} = 3$ for $0 ° ⩽ θ ⩽ 360 °$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5>Q 16

Show that the equation $\frac{4 \cos θ}{\tan θ} + 15 = 0$ can be expressed as $4 \sin^{2} θ - 15 \sin θ - 4 = 0$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5>Q 16

Hence, solve the equation $\frac{4 \cos θ}{\tan θ} + 15 = 0$ for $0 ° \leq θ \leq 360 °$ .

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>END-OF-CHAPTER REVIEW EXERCISE 5>Q 17

The function $f : x \to 5 + 3 \cos (\frac{1}{2} x)$ is defined for $0 \leq x ⩽ 2 π$ . Solve the equation $f (x) = 7$ , giving your answer correct to $2$ decimal places.