Moderate -0.3 This is a standard C4 harmonic form question with routine application of R cos α and R sin α formulas, followed by straightforward maximization. The technique is well-practiced and requires no novel insight—just methodical expansion, comparison of coefficients, and recognition that maximum occurs when sin(x+α)=1. Slightly easier than average due to being a textbook-standard application.
Express \(3\sin x + 2\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\)
Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = f(x)\), where
$$f(x) = 3\sin x + 2\cos x, \quad 0 \leqslant x \leqslant \pi.$$ [7]
3sin x +2cos x = R sin(x + α) = R sin x cos α + R cos x sin α
R cos α = 3, R sin α = 2
R2 = 32 + 22 = 13, R = 13
tan α = 2/3,
α = 0.588
3sinx2cosx 13sin(x0.588)
maximum when x + 0.588 = /2
x = /2 0.588 = 0.98 rads
y = 13 = 3.61
Answer
Marks
So coords of max point are (0.98, 3.61)
M1
B1
M1
A1
M1
A1
B1
Answer
Marks
[7]
Correct pairs. Condone omission of R if used correctly. Condone
sign error.
or 3.6 or better, not ±√13 unless+√13 chosen
ft from first M1
0.588 or better (accept 0.59), with no errors seen in method for
angle (allow 33.7º or better)
any valid method eg differentiating
0.98 only. Do not accept degrees or multiples of π.
condone √13, ft their R if ,say =√14
2sincos
3 LHS =
12cos21
2sincos
2cos2
sin
tanRHS
Answer
Marks
cos
M1
M1
E1
Answer
Marks
[3]
one correct double angle formula used
cancelling cos ’s
4(i) sin(θ + 45°) = cos θ
⇒ sin θ cos 45 + cos θ sin 45 = cos θ
⇒ (1/√2) sin θ + (1/√2) cos θ = cos θ
⇒ sin θ + cos θ = √2 cos θ
⇒ sin θ = (√2 – 1) cos θ
sinθ
⇒ = tan θ = √2 – 1 *
Answer
Marks
cosθ
M1
B1
A1
M1
E1
Answer
Marks
[5]
compound angle formula
sin 45 = 1/√2, cos 45 = 1/√2
collecting terms
(ii) tan θ = √2 – 1
⇒ θ = 22.5°,
Answer
Marks
202.5°
B1
B1
Answer
Marks
[2]
and no others in the range
5(i) 2sin 2θ + cos 2θ = 1
⇒ 4sin θ cos θ + 1 − 2sin2 θ = 1
⇒ 2sin θ (2cos θ − sin θ) = 0 or 4
tanθ-2tan²θ=0
⇒ sin θ = 0 or tan θ =0, θ = 0°, 180°
orro 2c θ − sin θ = 0
⇒ tan θ = 2
⇒ θ = 63.43°, 243.43°
OR
Using Rsin(2θ+α)
R=√5 and α=26.57° 2θ
+26.57=arcsin 1/R
θ=0° ,180°
Answer
Marks
θ=63.43°, 243.43°
M1
A1
A1
M1
A1,
A1
[6]
M1
A1
M1
A1
A1,A1
Answer
Marks
[6]
Using double angle formulae Correct
simplification to factorisable or other form
that leads to solutions 0° and 180°
tan θ = 2
(-1 for extra solutions in range)
(-1 for extra solutions in range)
Question 2:
2 | 3sin x +2cos x = R sin(x + α) = R sin x cos α + R cos x sin α
R cos α = 3, R sin α = 2
R2 = 32 + 22 = 13, R = 13
tan α = 2/3,
α = 0.588
3sinx2cosx 13sin(x0.588)
maximum when x + 0.588 = /2
x = /2 0.588 = 0.98 rads
y = 13 = 3.61
So coords of max point are (0.98, 3.61) | M1
B1
M1
A1
M1
A1
B1
[7] | Correct pairs. Condone omission of R if used correctly. Condone
sign error.
or 3.6 or better, not ±√13 unless+√13 chosen
ft from first M1
0.588 or better (accept 0.59), with no errors seen in method for
angle (allow 33.7º or better)
any valid method eg differentiating
0.98 only. Do not accept degrees or multiples of π.
condone √13, ft their R if ,say =√14
2sincos
3 LHS =
12cos21
2sincos
2cos2
sin
tanRHS
cos | M1
M1
E1
[3] | one correct double angle formula used
cancelling cos ’s
4(i) sin(θ + 45°) = cos θ
⇒ sin θ cos 45 + cos θ sin 45 = cos θ
⇒ (1/√2) sin θ + (1/√2) cos θ = cos θ
⇒ sin θ + cos θ = √2 cos θ
⇒ sin θ = (√2 – 1) cos θ
sinθ
⇒ = tan θ = √2 – 1 *
cosθ | M1
B1
A1
M1
E1
[5] | compound angle formula
sin 45 = 1/√2, cos 45 = 1/√2
collecting terms
(ii) tan θ = √2 – 1
⇒ θ = 22.5°,
202.5° | B1
B1
[2] | and no others in the range
5(i) 2sin 2θ + cos 2θ = 1
⇒ 4sin θ cos θ + 1 − 2sin2 θ = 1
⇒ 2sin θ (2cos θ − sin θ) = 0 or 4
tanθ-2tan²θ=0
⇒ sin θ = 0 or tan θ =0, θ = 0°, 180°
orro 2c θ − sin θ = 0
⇒ tan θ = 2
⇒ θ = 63.43°, 243.43°
OR
Using Rsin(2θ+α)
R=√5 and α=26.57° 2θ
+26.57=arcsin 1/R
θ=0° ,180°
θ=63.43°, 243.43° | M1
A1
A1
M1
A1,
A1
[6]
M1
A1
M1
A1
A1,A1
[6] | Using double angle formulae Correct
simplification to factorisable or other form
that leads to solutions 0° and 180°
tan θ = 2
(-1 for extra solutions in range)
(-1 for extra solutions in range)
Express $3\sin x + 2\cos x$ in the form $R\sin(x + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$
Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve $y = f(x)$, where
$$f(x) = 3\sin x + 2\cos x, \quad 0 \leqslant x \leqslant \pi.$$ [7]
\hfill \mbox{\textit{OCR C4 Q2 [7]}}