Date | May Specimen paper | Marks available | 7 | Reference code | SPM.2.AHL.TZ0.12 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Verify | Question number | 12 | Adapted from | N/A |
Question
Show that cot2θ=1−tan2θ2tanθ.
Verify that x=tanθ and x=−cotθ satisfy the equation x2+(2cot2θ)x−1=0.
Hence, or otherwise, show that the exact value of tanπ12=2−√3.
Using the results from parts (b) and (c) find the exact value of tanπ24−cotπ24.
Give your answer in the form a+b√3 where a, b∈Z.
Markscheme
stating the relationship between cot and tan and stating the identity for tan2θ M1
cot2θ=1tan2θ and tan2θ=2tanθ1−tan2θ
⇒ cot2θ=1−tan2θ2tanθ AG
[1 mark]
METHOD 1
attempting to substitute tanθ for x and using the result from (a) M1
LHS = tan2θ+2tanθ(1−tan2θ2tanθ)−1 A1
tan2θ+1−tan2θ−1=0(= RHS) A1
so x=tanθ satisfies the equation AG
attempting to substitute −cotθ for x and using the result from (a) M1
LHS = cot2θ−2cotθ(1−tan2θ2tanθ)−1 A1
=1tan2θ−(1−tan2θtan2θ)−1 A1
1tan2θ−1tan2θ+1−1=0(= RHS) A1
so x=−cotθ satisfies the equation AG
METHOD 2
let α=tanθ and β=−cotθ
attempting to find the sum of roots M1
α+β=tanθ−1tanθ
=tan2θ−1tanθ A1
=−2cot2θ (from part (a)) A1
attempting to find the product of roots M1
αβ=tanθ×(−cotθ) A1
= −1 A1
the coefficient of x and the constant term in the quadratic are 2cot2θ and −1 respectively R1
hence the two roots are α=tanθ and β=−cotθ AG
[7 marks]
METHOD 1
x=tanπ12 and x=−cotπ12 are roots of x2+(2cotπ6)x−1=0 R1
Note: Award R1 if only x=tanπ12 is stated as a root of x2+(2cotπ6)x−1=0.
x2+2√3x−1=0 A1
attempting to solve their quadratic equation M1
x=−√3±2 A1
tanπ12>0 (−cotπ12<0) R1
so tanπ12=2−√3 AG
METHOD 2
attempting to substitute θ=π12 into the identity for tan2θ M1
tanπ6=2tanπ121−tan2π12
tan2π12+2√3tanπ12−1=0 A1
attempting to solve their quadratic equation M1
tanπ12=−√3±2 A1
tanπ12>0 R1
so tanπ12=2−√3 AG
[5 marks]
tanπ24−cotπ24 is the sum of the roots of x2+(2cotπ12)x−1=0 R1
tanπ24−cotπ24=−2cotπ12 A1
=−22−√3 A1
attempting to rationalise their denominator (M1)
=−4−2√3 A1A1
[6 marks]