Loading [MathJax]/jax/element/mml/optable/Latin1Supplement.js

User interface language: English | Español

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=1tan2θ2tanθ.

[1]
a.

Verify that x=tanθ and x=cotθ satisfy the equation x2+(2cot2θ)x1=0.

[7]
b.

Hence, or otherwise, show that the exact value of tanπ12=23.

[5]
c.

Using the results from parts (b) and (c) find the exact value of tanπ24cotπ24.

Give your answer in the form a+b3 where abZ.

[6]
d.

Markscheme

stating the relationship between cot and tan and stating the identity for tan2θ       M1

cot2θ=1tan2θ and tan2θ=2tanθ1tan2θ

⇒ cot2θ=1tan2θ2tanθ     AG

[1 mark]

a.

METHOD 1

attempting to substitute tanθ for x and using the result from (a)      M1

LHS = tan2θ+2tanθ(1tan2θ2tanθ)1      A1

tan2θ+1tan2θ1=0(= RHS)      A1

so x=tanθ satisfies the equation      AG

attempting to substitute cotθ for x and using the result from (a)       M1

LHS = cot2θ2cotθ(1tan2θ2tanθ)1      A1

=1tan2θ(1tan2θtan2θ)1      A1

1tan2θ1tan2θ+11=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=tanθ1tanθ

         =tan2θ1tanθ     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]

b.

METHOD 1

x=tanπ12 and x=cotπ12 are roots of x2+(2cotπ6)x1=0        R1

Note: Award R1 if only x=tanπ12 is stated as a root of x2+(2cotπ6)x1=0.

x2+23x1=0        A1

attempting to solve their quadratic equation         M1

x=3±2        A1

tanπ12>0  (cotπ12<0)        R1

so tanπ12=23      AG

 

METHOD 2

attempting to substitute θ=π12 into the identity for tan2θ           M1

tanπ6=2tanπ121tan2π12

tan2π12+23tanπ121=0     A1

attempting to solve their quadratic equation      M1

tanπ12=3±2     A1

tanπ12>0      R1

so tanπ12=23      AG

[5 marks]

c.

tanπ24cotπ24 is the sum of the roots of x2+(2cotπ12)x1=0        R1

tanπ24cotπ24=2cotπ12      A1

=223      A1

attempting to rationalise their denominator       (M1)

=423       A1A1

[6 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots
Show 47 related questions
Topic 3— Geometry and trigonometry » AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions
Topic 3— Geometry and trigonometry » AHL 3.10—Compound angle identities
Topic 2—Functions
Topic 3— Geometry and trigonometry

View options