Maths Extension 1 Year 11 - Module 3 - Lesson 15
Module Synthesis & Exam Technique
Name
Date
Class
1. Lesson focus
Mixed trigonometric synthesis. Choose an identity or transformation that reduces the equation, solve completely in the specified interval, and check excluded values.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. Solve 2sin²x−1=0 for 0≤x<2π.
3. Solve 3sinx+4cosx=5 for 0≤x<2π.
3. Challenge
Show that (sinx+cosx)²=1+sin2x, then state its maximum.
4. Multiple choice
Which expression equals 1+sin2x?
A. (sinx+cosx)²
B. (sinx−cosx)²
C. sin²x−cos²x
D. 1−2sinxcosx
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. Simplify sin²x+cos²x+tan²x−sec²x where defined.
Using sin²x+cos²x=1 and sec²x−tan²x=1, the expression is 0 wherever defined. Since tan x and sec x require cos x≠0, exclude x=π/2+kπ for integer k.
2. Solve 2sin²x−1=0 for 0≤x<2π.
sin²x=1/2, so sinx=±√2/2; x=π/4,3π/4,5π/4,7π/4.
3. Solve 3sinx+4cosx=5 for 0≤x<2π.
5sin(x+α)=5 where α=arctan(4/3); hence x+α=π/2 mod 2π, so x=π/2−α.
Challenge. Show that (sinx+cosx)²=1+sin2x, then state its maximum.
Expand: sin²+cos²+2sinxcosx=1+sin2x. Since sin2x≤1, maximum is 2.
Multiple choice. Which expression equals 1+sin2x?
Correct answer: A. Expand the square and use sin²x+cos²x=1 and sin2x=2sinxcosx.
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