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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  cot 2 θ = 1 ta n 2 θ 2 tan θ .

[1]
a.

Verify that  x = tan θ and  x = cot θ satisfy the equation x 2 + ( 2 cot 2 θ ) x 1 = 0 .

[7]
b.

Hence, or otherwise, show that the exact value of  tan π 12 = 2 3 .

[5]
c.

Using the results from parts (b) and (c) find the exact value of  tan π 24 cot π 24 .

Give your answer in the form  a + b 3 where  a b Z .

[6]
d.

Markscheme

stating the relationship between cot and tan and stating the identity for tan 2 θ        M1

cot 2 θ = 1 tan 2 θ and  tan 2 θ = 2 tan θ 1 ta n 2 θ

⇒  cot 2 θ = 1 ta n 2 θ 2 tan θ      AG

[1 mark]

a.

METHOD 1

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

LHS =  ta n 2 θ + 2 tan θ ( 1 ta n 2 θ 2 tan θ ) 1       A1

ta n 2 θ + 1 ta n 2 θ 1 = 0 (= RHS)      A1

so  x = tan θ  satisfies the equation      AG

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

LHS =  co t 2 θ 2 cot θ ( 1 ta n 2 θ 2 tan θ ) 1       A1

= 1 ta n 2 θ ( 1 ta n 2 θ ta n 2 θ ) 1       A1

1 ta n 2 θ 1 ta n 2 θ + 1 1 = 0 (= RHS)     A1

so  x = cot θ satisfies the equation      AG

 

METHOD 2

let  α = tan θ and  β = cot θ

attempting to find the sum of roots       M1

α + β = tan θ 1 tan θ

          = ta n 2 θ 1 tan θ      A1

          = 2 cot 2 θ (from part (a))     A1

attempting to find the product of roots         M1

α β = tan θ × ( cot θ )      A1

= −1     A1

the coefficient of x and the constant term in the quadratic are 2 cot 2 θ and −1 respectively        R1

hence the two roots are  α = tan θ  and  β = cot θ        AG

[7 marks]

b.

METHOD 1

x = tan π 12 and  x = cot π 12 are roots of  x 2 + ( 2 cot π 6 ) x 1 = 0         R1

Note: Award R1 if only  x = tan π 12  is stated as a root of  x 2 + ( 2 cot π 6 ) x 1 = 0 .

x 2 + 2 3 x 1 = 0         A1

attempting to solve their quadratic equation         M1

x = 3 ± 2         A1

tan π 12 > 0   ( cot π 12 < 0 )        R1

so  tan π 12 = 2 3       AG

 

METHOD 2

attempting to substitute  θ = π 12 into the identity for  tan 2 θ            M1

tan π 6 = 2 tan π 12 1 ta n 2 π 12

ta n 2 π 12 + 2 3 tan π 12 1 = 0      A1

attempting to solve their quadratic equation      M1

tan π 12 = 3 ± 2      A1

tan π 12 > 0       R1

so  tan π 12 = 2 3       AG

[5 marks]

c.

tan π 24 cot π 24  is the sum of the roots of  x 2 + ( 2 cot π 12 ) x 1 = 0         R1

tan π 24 cot π 24 = 2 cot π 12       A1

= 2 2 3       A1

attempting to rationalise their denominator       (M1)

= 4 2 3        A1A1

[6 marks]

d.

Examiners report

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

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots
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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

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