Given that cos θ = - 1 2 , and θ is obtuse, find: tan 2 θ

Given, cos θ = - 1 2 , and θ is obtuse. We need to find the value of tan 2 θ . We know that cos 2 θ = 2 cos 2 θ - 1 = 2 - 1 2 2 - 1 = 2 1 4 - 1 = 1 2 - 1 = - 1 2 . . . . i Now, we know that sin 2 θ + cos 2 θ = 1 ⇒ sin 2 θ = 1 - cos 2 θ = 1 - - 1 2 2 = 1 - 1 4 = 3 4 ⇒ sin θ = 3 2 . . . . i i [ ∵ sin θ is positive if θ lies in the second quadrant ] Now, we know that sin 2 θ = 2 sin θ cos θ = 2 × 3 2 × - 1 2 [using i i ] = - 3 2 . . . . . i i i Now, we know that tan 2 θ = sin 2 θ cos 2 θ = - 3 2 - 1 2 = 3 Hence, tan 2 θ = 3

Given that sin θ = - 1 8 , and π ≤ θ ≤ 3 π 2 , find: cos 2 θ

Given, sin θ = - 1 8 , and π ≤ θ ≤ 3 π 2 . To find: cos 2 θ We know that cos 2 θ = 1 - 2 sin 2 θ = 1 - 2 - 1 8 2 = 1 - 2 1 64 = 1 - 1 32 = 31 32 Hence, cos 2 θ = 31 32

Solve the given equation for 0 ≤ θ ≤ 2 π giving your answer in terms of π . sin 2 θ = 3 cos θ

Given equation is sin 2 θ = 3 cos θ ⇒ 2 sin θ cos θ = 3 cos θ ⇒ cos θ 3 - 2 sin θ = 0 ⇒ cos θ = 0 or sin θ = 3 2 Consider cos θ = 0 ⇒ θ = π 2 , 3 π 2 . . . . i Now consider sin θ = 3 2 ⇒ sin θ = sin π 3 ⇒ θ = π 3 . . . . . i i Also, sin θ = sin π 3 = sin π - π 3 = sin 2 π 3 ⇒ θ = 2 π 3 . . . . i i i From i , i i , and i i i , we have θ = π 2 , 3 π 2 , π 3 , 2 π 3

Solve the given equation for 0 ≤ θ ≤ 2 π giving your answer in terms of π . cos θ = sin θ sin 2 θ

Given equation is cos θ = sin θ sin 2 θ ⇒ cos θ = sin θ 2 sin θ cos θ ⇒ 2 sin 2 θ cos θ - cos θ = 0 ⇒ cos θ 2 sin 2 θ - 1 = 0 ⇒ cos θ = 0 . . . . i Or 2 sin 2 θ = 1 ⇒ sin 2 θ = 1 2 ⇒ sin θ = 1 2 . . . . i i or sin θ = - 1 2 . . . . i i i From equation i , we have cos θ = 0 ⇒ θ = π 2 , 3 π 2 . . . . . iv From equation i i , we have sin θ = 1 2 = sin π 4 ⇒ θ = π 4 . . . . v And, sin θ = sin π 4 = sin π - π 4 = sin 3 π 4 ⇒ θ = 3 π 4 . . . . v i From equation i i i , we have sin θ = - 1 2 = - sin π 4 = sin π + π 4 = sin 5 π 4 ⇒ θ = 5 π 4 . . . . . v i i Also, sin θ = - sin π 4 = sin 2 π - π 4 = sin 7 π 4 ⇒ θ = 7 π 4 . . . . v i i i From equations, i v , v , v i , v i i , and v i i i , The solution of the given trigonometric equation is π 4 , π 2 , 3 π 4 , 5 π 4 , 3 π 2 , 7 π 4 .

Solve the given equation for 0 ≤ θ ≤ 2 π giving your answer in terms of π . cos 2 θ = cos θ

Given equation is cos 2 θ = cos θ ⇒ 2 cos 2 θ - 1 - cos θ = 0 ⇒ 2 cos 2 θ - cos θ - 1 = 0 Let cos θ = x ⇒ 2 x 2 - x - 1 = 0 ⇒ x = 1 ± - 1 2 - 4 × 2 × - 1 2 × 2 = 1 ± 9 4 = 1 ± 3 4 ⇒ x = 1 + 3 4 = 1 or x = 1 - 3 4 = - 2 4 = - 1 2 ⇒ cos θ = 1 or cos θ = - 1 2 Consider cos θ = 1 ⇒ θ = 0 , 2 π . . . . i Now consider cos θ = - 1 2 = - cos π 3 = cos π - π 3 = cos 2 π 3 ⇒ θ = 2 π 3 . . . . i i Also, cos θ = - cos π 3 = cos π + π 3 = cos 4 π 3 ⇒ θ = 4 π 3 . . . . i i i From i , i i , and i i i The solutions of the given trigonometric equation such that 0 ≤ θ ≤ 2 π is 0 , 2 π 3 , 4 π 3 , 2 π .

Given that triangle A B C has an area of 10 cm 2 when sin θ = 1 4 , find x .

Given, Area of △ A B C = 10 cm 2 , and sin θ = 1 4 To find: x . From the figure, We know that Area of △ A B C = 1 2 × 15 × x × sin 2 θ . . . . i We know that cos 2 θ = 1 - sin 2 θ = 1 - 1 4 2 = 1 - 1 16 = 15 16 ⇒ cos θ = 15 4 Now, sin 2 θ = 2 sin θ cos θ = 2 × 1 4 × 15 4 = 15 8 . . . . i i From i , we have △ A B C = 1 2 × 15 × x × 15 8 ⇒ 10 = 15 x 16 ⇒ x = 160 15 = 32 3 Hence, x = 32 3 cm

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F

Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 17: Exercise 12F

Author:Natasha Awada, Paul La Rondie, Laurie Buchanan & Jill Stevens

Natasha Awada Mathematics Solutions for Exercise - Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 17: Exercise 12F

Attempt the practice questions on Chapter 12: Periodic Relationships: Trigonometric Functions, Exercise 17: Exercise 12F with hints and solutions to strengthen your understanding. Mathematics : Analysis and Approaches Standard Level Course Companion solutions are prepared by Experienced Embibe Experts.

Questions from Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 17: Exercise 12F with Hints & Solutions

MEDIUM

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 3

Given that $\cos θ = - \frac{1}{2}$ , and $θ$ is obtuse, find:

$\tan 2 θ$

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 4

Given that $\sin θ = - \frac{1}{8}$ , and $π \leq θ \leq \frac{3 π}{2}$ , find:

$\cos 2 θ$

MEDIUM

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 5

Solve the given equation for $0 \leq θ \leq 2 π$ giving your answer in terms of $π$ .

$\sin 2 θ = \sqrt{3} \cos θ$

MEDIUM

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 5

Solve the given equation for $0 \leq θ \leq 2 π$ giving your answer in terms of $π$ .

$\cos θ = \sin θ \sin 2 θ$

MEDIUM

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 5

Solve the given equation for $0 \leq θ \leq 2 π$ giving your answer in terms of $π$ .

$\cos 2 θ = \cos θ$

EASY

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 6

The expression $32 \sin x \cos x$ can be expressed in the form $a \sin b x$ . Find the value of $a$ and of $b$ .

MEDIUM

Diploma

IMPORTANT

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12F>Q 6

The expression $32 \sin x \cos x$ can be expressed in the form $a \sin b x$ . Hence, solve the equation $32 \sin x \cos x = 8$ , for $0 \leq x \leq π$ .