C4 Trigonomnetry - Trigonometrical identities
1.
(a)
Attempt at integration by parts, i.e. kx sin 2x ± ∫k sin 2xdx,
with k = 2 or ½
1
1
= x sin 2x –
sin 2x dx
2
2
∫
PhysicsAndMathsTutor.com
M1
A1
1
1
x sin 2x + cos 2x + c
M1, A1
2
4
(penalise lack of constant of integration first time only)
Integrates sin 2x correctly, to obtain
(b)
B1
Hence method: Uses cos 2x = 2cos2 x – 1 to connect integrals
Obtains
1
1 x2
x2 x
x cos 2 xdx = { + answer to part(a )} =
+ sin 2 x + cos 2 x + k M1A1
8
2 2
4 4
Otherwise method
x
x
1
1
x cos 2 xdx = x sin 2 x +  −
sin 2 x + dx
B1, M1
2
4
2
4
∫
∫
4
3
∫
B1 for (
1
x
sin2x + )
4
2
x2 x
1
+ sin 2 x + cos 2 x + k
8
4 4
=
A1
3
[10]
2.
(a)
(b)
sin(3x + x) = sin3x cosx + cos3x sinx
sin(3x – x) = sin3x cosx – cos 3x sinx
(subtract) ⇒ sin4x – sin2a = 2sinx cos3x
∫2sinx cos3x dx = ∫(sin4x –sin2x)dx
cos 4 x cos 2 x
= −
+
+c
4
2
M1
A1
A1c.s.o.
3
M1
A1ft
2
∫their p, q
(c)
5π
6
∫π
2
=
1
5π   1
10π 1
 1

+ cos
2 sin x cos 3 xdx =  − cos
 −  − cos 2π + cos π  M1
2
3   4
3
2
 4

9
8
A1
2
[7]
Edexcel Internal Review
3
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C4 Trigonomnetry - Trigonometrical identities
3.
(a)
(b)
cos(A + A) = cos2 A – sin2 A
= cos2 A – (1 – cos2 A) = 2 cos2 A – 1
M1
A1
π
π
;x= 6,θ=
]
4
3
dx
= 2 2 cos θ
x = 2 2 sinθ,
dθ
2
2
∫ 8 − x dx = ∫ 2 2 cosθ 2 2 cosθdθ = ∫ 8 cos θdθ
[x = 2, θ =
Using cos 2θ = 2 cos2 θ – 1 to give
∫ 4(1 + cos 2θ )dθ
= 4θ + 2 sin 2θ
Substituting limits to give
(c)
1
π + 3 − 2 or given result
3
dy − 2 sin 2θ
=
dθ 1 + cos 2θ
2
B1
B1
M1 A1
dM1
A1 ft
A1
7
B1
dy
dy
= sec θ tan θ to give
(= –2 cos θ)
dθ
dx
Gradient at the point where θ = π3 is –1.
Using the chain rule, with
Equation of tangent is y + ln 2 =
−( x − 2) (o.a.e.)
M1
A1 ft
M1 A1
5
[12]
4.
(a)
y
(1, 2 )
2
–1
x
O
–2
(1, – 2 )
y = arcsin x
(a)
Shape correct
passing through O:
end-points:
Edexcel Internal Review
G1;
G1
2
4
PhysicsAndMathsTutor.com
C4 Trigonomnetry - Trigonometrical identities
y
( 3 , 2)
(– 3 , 2)
O
x
(b)
y = sec x
Shape correct,
symmetry in Oy:
end-points:
(c)
x
−
sec x
2
π
3
Area estimate =
= 2.78 (2 d.p.)
G1
G1
−
π
6
1.155
∫
0
1
π
3
π
−3
sec x dx =
π
6
1.155
π 2 + 2
2
π
3
2

+ 1.155 + 1 + 1.155

6 2

M1 A1 A1
A1
4
[8]
Edexcel Internal Review
5
C4 Trigonomnetry - Trigonometrical identities
1.
PhysicsAndMathsTutor.com
(a)
This was a straightforward integration by parts, which was recognised as such and done
well in general. The most common error was the omission of the constant of integration,
but some confused signs and others ignored the factors of two.
(b)
This was done well by those students who recognised that cos2x = (1 + cos2x)/2 but there
was a surprisingly high proportion who were unable to begin this part. Lack of care with
brackets often led to errors so full marks were rare. There was also a large proportion of
candidates who preferred to do the integration by parts again rather than using their
answer to (a).
2.
Candidates commonly misunderstood that both the formulae for sin(A+B) and sin(A–B) were
required by the rubric. The vast majority chose only one, prohibiting progress and usually
abandoned the question at this stage. Many who did not complete part (a) attempted to integrate
by parts, usually twice, in part (b) before leaving unfinished working. The more successful, or
those with initiative, continued with both parts (b) and (c), either with their p and q values, the
letters p and q, or hopefully guessed p and q values.
3.
Most candidates understood the requirements of the proof of the double angle formula in part
(a). Part (b) proved to be discriminating, but a large number of candidates produced good
solutions, where they changed the variables and the limits and used the appropriate double angle
formula to perform the integration. Some difficulties were experienced differentiating the log
function in part (c), but again there were a large number of correct solutions. A few candidates
eliminated the parameter and found the cartesian equation of the curve before differentiation.
4. No Report available for this question.
Edexcel Internal Review
6