Edexcel Past-Year Question (Q10 P1, 28/5/2004)
2)
sin(A + B) ≡ sin A cos B + cos A sin
B
By expanding R sin(x + Ө),
(a)
Find, in degrees to 1 decimal place, the value of Ө
such that
5 sin x + 12 cos x ≡ R sin(x + Ө)
Where R > 0, 0 < Ө < 90o and 0 < x < 360o. (5)
(b)
Show that R = 13. (2)
(c)
Write down the minimum value of 5 sin x + 12 cos x. (1)
(d)
Find, to 1 decimal place, the value of x which gives
this minimum value. (2)
(e)
Solve the equation 5 sin x + 12 cos x = 8. (4)
My Proposed Solutions
2
(a) Expanding, R sin(x + Ө) = R [sin x cos Ө + cos x sin Ө]
= R sin
x cos Ө + R cos x sin Ө
≡ 5 sin x + 12 cos x
Comparing coefficients: R cos Ө = 5 …Eqn
(1)
R sin Ө = 12 …Eqn (2)
(2)
÷ (1): (R sin
Ө)/( R cos Ө) = tan Ө = 12/5
tan Ө = 2.4
For
domain 0 < Ө < 90o: Ө =
tan-1 2.4 = 67.38…o
Answer:
Ө = 67.4o (1 d.p.)
2
(b) Using eqn (1): R =
5/cos Ө = 5/cos (tan-1 2.4) = 13 (Shown)
2
(c) Since, 5 sin x + 12 cos
x ≡ R sin(x + Ө)
Implying: Min. value occurs when R sin(x + Ө) is minimum, and
R sin(x + Ө)
is minimum when sin(x + Ө) = - 1
Therefore: Min. value of 5 sin x + 12 cos x ≡ 13(-1) = - 13 (Answer)
2
(d) To find x which gives the minimum
value:
Use: R
sin(x + Ө) = - 13, where R = 13 and Ө = tan-1 2.4
sin(x + Ө) = - 1
Domain for (x + Ө): 0 + 67.38…o < (x + Ө) <
360o + 67.38…o
Therefore: (x + Ө) = sin-1(-1)
= 270o
Hence, x = 270o – Ө = 270o
- tan-1 2.4 = 202.61…o
Answer: x = 202.6o (1 d.p.)
2
(e) To solve: 5 sin x +
12 cos x = 8
Use: R sin(x + Ө) = 8, where R = 13 and Ө
= tan-1 2.4
Therefore: sin(x + Ө)
= 8/13
Domain for (x + Ө): 0 + 67.38…o < (x + Ө) <
360o + 67.38…o
Reference Angle, R: R = sin-1 (8/13) = 37.97…o
(< 67.38…o)
Therefore, (x + Ө) = 180o
- R (2nd Qd), or, = 360o + R (5th Qd)
Hence, x = 180o – R – Ө or, = 360o + R - Ө
Thus, x
= 180o - sin-1 (8/13) - tan-1 2.4 = 74.63…o
or, x = 360o + sin-1
(8/13) - tan-1 2.4 = 330.59…o
Answer: x = 74.6o
or 330.6o (1 d.p.)
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