By first expanding (cos θ + sin θ)², find the three solutions of the equation (cos θ + sin θ)² = 1 for 0 ≤ θ < π.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [*B1] cos²θ + 2sinθcosθ + sin²θ = 1 leading to 2sinθcosθ = 0 or sin 2θ = 0 Or arriving at cosθ = 0 or sin θ = 0 or tanθ = 0 after first expanding and www.
2. [DB2,1,0] [θ=] 0, π/2, π B2 for three correct answers only. B1 for two correct answers and one incorrect or 3 correct answers plus other values in the range. SC DB1 for correct 3 answers in degrees and no others. Ignore extras outside of the range and allow decimal equivalents.
3. [B1] cos0+sin0=[1+0=]1 and cos(π/2)+sin(π/2)[=0+1]=1 Checking both correct values. Do not allow solving an equation. Condone use of 90 degrees.
4. [B1] cosπ+sinπ[=−1+0]=−1 or ≠1 www
5. [M1] (cosθ-sinθ)sinθ + (cosθ+sinθ)(1-cosθ) / (cosθ+sinθ)(cosθ-sinθ) Correct common denominator and correct products in the numerator and no missing terms. Correct factors in the denominator can be implied by cos²θ – sin²θ . Condone brackets missing if recovered.
6. [A1] cosθsinθ-sin²θ + cosθ-cos²θ+sinθ-sinθcosθ / cos²θ-sin²θ
7. [A1] sinθ + cosθ - cos²θ - sin²θ / cos²θ-sin²θ = cosθ+sinθ-1 / 1-2sin²θ AG Clear evidence of using sin²θ+cos²θ=1 in either the numerator or denominator. Condone c, s and/or omission of θ. Working from both sides of the identity and correctly arriving at the same expression can score M1A1. A final statement is then required for the A1.
8. [*M1] Replacing LHS with the expression from (b) and attempting to simplify i.e. condone omission of (cosθ+sinθ-1)=0 at this stage. M0 for 0 = 2(1-2sin²θ)
9. [DM1] Dividing by k and taking the square root of a positive value <1.
10. [A1] Solutions 0, π/6, 5π/6, π/2. Allow 0, 0.524, 1.57, 2.62 AWRT. SC: If M0 SCB1 for (cosθ+sinθ-1)=0 => θ=0, π/2. SC: If M0 SCB1 for all four correct answers and no others. Ignore answers outside of the range. Answers in degrees A0.