Prove the identity cos θ tan θ ( 1 + sin θ ) = 1 sin θ - 1

Given that, cos θ tan θ ( 1 + sin θ ) ≡ 1 sin θ - 1 Left-hand side: Multiplying top and bottom by 1 - sin θ ⇒ L . H . S = cos θ ( 1 - sin θ ) tan θ ( 1 + sin θ ) ( 1 - sin θ ) Expanding denominator: ⇒ L . H . S = cos θ ( 1 - sin θ ) tan θ 1 - sin 2 θ We know that, sin 2 θ + cos 2 θ ≡ 1 . ⇒ L . H . S = cos θ ( 1 - sin θ ) tan θ cos 2 θ Dividing top and bottom by cos θ gives: ⇒ L . H . S = 1 - sin θ sin θ cos θ × cos θ ⇒ L . H . S = 1 - sin θ sin θ ⇒ L . H . S = 1 sin θ - 1 ⇒ L . H . S = 1 sin θ - 1 = R . H . S ∴ L . H . S = R . H . S Hence proved,

Hence, solve the equation cos θ tan θ ( 1 + sin θ ) = 1 for 0 ° ≤ θ ≤ 360 °

Given that, cos θ tan θ ( 1 + sin θ ) = 1 , for 0 ° ≤ θ ≤ 360 ° ⇒ cos θ sin θ cos θ ( 1 + sin θ ) = 1 ⇒ cos 2 θ ( sin θ + sin 2 θ ) = 1 ⇒ cos 2 θ ( sin θ + sin 2 θ ) = 1 ⇒ 1 - sin 2 θ sin θ ( 1 + sin θ ) = 1 ( ∵ sin 2 θ + cos 2 θ = 1 ) ⇒ 1 - sin θ 1 + sin θ sin θ ( 1 + sin θ ) = 1 ⇒ 1 - sin θ sin θ = 1 ∴ 1 - sin θ = sin θ ∴ 1 = 2 sin θ ∴ sin θ = 1 2 Now, sin θ = 1 2 ⇒ sin θ = sin π 6 ⇒ θ = π 6 Also, sin θ = 1 2 ⇒ sin θ = sin 5 π 6 ⇒ θ = 5 π 6

Prove the identity 1 1 + sin θ + 1 1 - sin θ = 2 cos 2 θ

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

Hence, solve the equation cos θ 1 1 + sin θ + 1 1 - sin θ = 5 for 0 ° ≤ θ ≤ 360 °

Given that, cos θ 1 1 + sin θ + 1 1 - sin θ = 5 , 0 ° ≤ θ ≤ 360 ° ⇒ cos θ 1 1 + sin θ + 1 1 - sin θ = 5 ⇒ 1 1 + sin θ + 1 1 - sin θ = 1 - sin θ + 1 + sin θ 1 + sin 2 θ = 2 1 - sin 2 θ = 2 cos 2 θ ∵ sin 2 θ + cos 2 θ = 1 ⇒ cos θ × 2 cos 2 θ = 5 ⇒ 5 cos 2 θ = 2 cos θ ∴ 5 cos 2 θ - 2 cos θ = 0 ⇒ cos θ ( 5 cos θ - 2 ) = 0 ∴ cos θ = 0 or 5 cos θ - 2 = 0 ⇒ cos θ = 0 or cos θ = 2 5 Let, cos θ = 0 ⇒ θ = π 2 , 3 π 2 ⇒ θ = 90 ° , 270 ° Also, cos θ = 2 5 ∴ θ = cos - 1 2 5

Prove the identity 1 sin θ + 1 tan θ 2 = 1 + cos θ 1 - cos θ

Given that, 1 sin θ + 1 tan θ 2 ≡ 1 + cos θ 1 - cos θ Left-hand side: L . H . S = 1 sin θ + 1 tan θ 2 Using tan θ = sin θ cos θ and substituting gives: L . H . S = 1 sin θ + 1 sin θ cos θ 2 L . H . S = = 1 sin θ + cos θ sin θ 2 = 1 sin θ + cos θ sin θ 2 Adding fractions gives: = 1 + cos θ 2 sin 2 θ Squaring gives: L . H . S = ( 1 + cos θ ) ( 1 + cos θ ) sin 2 θ We know that, sin 2 θ + cos 2 θ ≡ 1 L . H . S = ( 1 + cos θ ) ( 1 + cos θ ) 1 - cos 2 θ Writing the denominator as the difference of two squares give: L . H . S = ( 1 + cos θ ) ( 1 + cos θ ) ( 1 - cos θ ) ( 1 + cos θ ) Dividing top and bottom by ( 1 + cos θ ) gives: L . H . S = 1 + cos θ 1 - cos θ = R . H . S L . H . S = R . H . S Hence proved,

Solve the equation 1 sin θ + 1 tan θ 2 = 2 for 0 ° ≤ θ ≤ 360 ° .

Given that, 1 sin θ + 1 tan θ 2 = 2 , 0 ° ≤ θ ⩽ 360 * ⇒ 1 sin θ + 1 tan θ 2 = 2 ⇒ 1 sin θ + 1 sin θ cos θ 2 = 2 ⇒ 1 sin θ + cos θ sin θ 2 = 2 ⇒ 1 + cos θ sin θ 2 = 2 We know that, sin 2 x + cos 2 x = 1 . ⇒ 1 + cos θ 2 1 - cos 2 θ = 2 ⇒ 1 + cos θ 2 1 - cos θ 1 + cos θ = 2 ⇒ 1 + cos θ 1 - cos θ = 2 1 + cos θ = 2 ( 1 - cos θ ) ⇒ 1 + cos θ = 2 - 2 cos θ ∴ 3 cos θ = 1 ∴ cos θ = 1 3 ∴ cos θ = 1 3 ⇒ θ = cos - 1 1 3 ∴ θ = 70 . 73 ° or 289 . 27 °

Hence, solve the equation cos 4 θ - sin 4 θ = 1 2 for 0 ° ≤ θ ≤ 360 °

Given that, cos 2 θ 2 - sin 2 θ 2 = 1 2 ⇒ cos 2 θ 2 - sin 2 θ 2 = 1 2 ⇒ cos 2 θ - sin 2 θ sin 2 θ + cos 2 θ = 1 2 ⇒ cos 2 θ - sin 2 θ = 1 2 We know that, sin 2 θ + cos 2 θ = 1 ⇒ cos 2 θ - 1 - cos 2 θ = 1 2 ⇒ cos 2 θ - 1 + cos 2 θ = 1 2 ⇒ 2 cos 2 θ - 1 = 1 2 ⇒ 2 cos 2 θ = 3 2 ⇒ cos 2 θ = 3 4 ⇒ cos 2 θ = 3 2 2 ∴ cos 2 θ = cos 2 π 6 We know that, cos 2 θ = cos 2 α then θ = n π ± α , where n ∈ z ⇒ θ = n π ± π 6 Where n ∈ z Put n = 1 , θ = π + π 6 , θ = π - π 6 Therefore the solution of the given equation is θ = 7 π 6 , θ = 5 π 6 Where, 0 ° < θ < 360 ° .

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 5 - Trigonometry>EXERCISE 5H

Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 11: EXERCISE 5H

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Trigonometry, Exercise 11: EXERCISE 5H

Attempt the practice questions on Chapter 5: Trigonometry, Exercise 11: EXERCISE 5H 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.