Trigonometry –
Addition Formulae:
Edexcel syllabus on
this segment reads:
“The use of the basic addition formulae of trigonometry.
Formal proofs of the basic formulae will not be required; questions using the
formulae for sin(A + B), cos(A + B), tan(A + B) may be set; the formulae will
be provided, for example:
sin(A + B)
= sin A cos B + cos A sin B
Long questions, explicitly involving excessive manipulation,
will not be set.”
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Edexcel Past Year
Questions Involving Addition Formulae:
[(27.12.2013): Try the questions first. Then only look for my proposed solutions either in the link given under the question or upon your email request. It's time consuming to get them all typed and posted. Nevertheless, I blog herewith the solution for Q1 to show you that these are all very interesting questions. Have fun and try them all!]
1)
sin (A + B) ≡ sin A cos B + cos A
sin B
sin (A - B) ≡
sin A cos B – cos A sin B
(a)
Show that sin (A + B) + sin (A - B) ≡ 2 sin A cos B. (1)
(b)
Hence show that sin P + sin Q ≡ 2 sin [(P+Q)/2] cos [(P-Q)/2]. (3)
(c)
Solve for x, 0 ≤ x ≤ π/2, giving your answers in terms of
π,
the equation sin 5x + sin 3x = 0. (5)
(d)
Show that sin 6x + 2 sin 4x + sin 2x ≡ 4 sin 4x cos2x. (4)
(e)
Hence…(Form 5 ‘Integration’) (24/3/2004 P1 Q11)
My Proposed Solution:
1(a) LHS = sin (A + B) + sin (A - B)
= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)
= 2 sin A cos B
= RHS (Shown)
1(b) Let: P = A + B; and Q = A - B
Therefore: P + Q = 2A; implying: A = (P+Q)/2
P - Q = 2B; implying: B = (P-Q)/2
Since: sin (A + B) + sin (A - B) ≡ 2 sin A cos B (see 1(a))
Imply: LHS = sin P + sin Q ≡ 2 sin A cos B = 2 sin [(P+Q)/2] cos [(P-Q)/2]
= RHS (Shown)
1(c) sin 5x + sin 3x = 0
Use 1(b): Therefore, 2 sin [(5x+3x)/2] cos [(5x-3x)/2] = 0
2 sin 4x cos x = 0
Imply: cos x = 0 or, sin 4x = 0 for 0 ≤ 4x ≤ 2π
x = π/2 or, 4x = 0, π or 2π
or, x = 0, π/4 or π/2
1(d) LHS = sin 6x + 2 sin 4x + sin 2x
= (sin 6x + sin 2x) + 2 sin 4x (since RHS also has term: sin 4x)
= 2 sin [(6x+2x)/2] cos [(6x-2x)/2] + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
From here, if you remember or are given the identity: cos 2x ≡ 2 cos2 x – 1
LHS = 2 sin 4x [(2 cos2 x – 1) + 1]
= 2 sin 4x (2 cos2 x)
= 4 sin 4x cos2x
= RHS (Shown)
Alternatively, use these relationships (if cos 2x ≡ 2 cos2 x – 1 is not given)
* cos 2x = sin (π/2 - 2x) ....Sine and cosine of complimentary angles are equal;
* 1 = sin π/2 ...............Trigo ratio of convenient angle: sin π/2 = sin 90o = 1
Thus, LHS = 2 sin 4x (cos 2x + 1)
= (2 sin 4x) [sin (π/2 - 2x) + sin π/2]
= (2 sin 4x){2 sin [(π/2 - 2x + π/2)/2] cos [(-2x)/2]}
= (4 sin 4x) sin (π/2 - x) cos (-x)
= (4 sin 4x) cos x cos x ..........[sin (π/2 - x) = cos x]
= 4 sin 4x cos2x
= RHS (Shown)
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If you ponder over Q1 above and its solutions, you should also be able to solve this:
Extra Q. Find the exact value of: sin 75o + sin 15o. (Answer: √(3/2))
P/s: If you don't know, email me and kindly introduce yourself when requesting something.
My Proposed Solution:
1(a) LHS = sin (A + B) + sin (A - B)
= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)
= 2 sin A cos B
= RHS (Shown)
1(b) Let: P = A + B; and Q = A - B
Therefore: P + Q = 2A; implying: A = (P+Q)/2
P - Q = 2B; implying: B = (P-Q)/2
Since: sin (A + B) + sin (A - B) ≡ 2 sin A cos B (see 1(a))
Imply: LHS = sin P + sin Q ≡ 2 sin A cos B = 2 sin [(P+Q)/2] cos [(P-Q)/2]
= RHS (Shown)
1(c) sin 5x + sin 3x = 0
Use 1(b): Therefore, 2 sin [(5x+3x)/2] cos [(5x-3x)/2] = 0
2 sin 4x cos x = 0
Imply: cos x = 0 or, sin 4x = 0 for 0 ≤ 4x ≤ 2π
x = π/2 or, 4x = 0, π or 2π
or, x = 0, π/4 or π/2
1(d) LHS = sin 6x + 2 sin 4x + sin 2x
= (sin 6x + sin 2x) + 2 sin 4x (since RHS also has term: sin 4x)
= 2 sin [(6x+2x)/2] cos [(6x-2x)/2] + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
From here, if you remember or are given the identity: cos 2x ≡ 2 cos2 x – 1
LHS = 2 sin 4x [(2 cos2 x – 1) + 1]
= 2 sin 4x (2 cos2 x)
= 4 sin 4x cos2x
= RHS (Shown)
Alternatively, use these relationships (if cos 2x ≡ 2 cos2 x – 1 is not given)
* cos 2x = sin (π/2 - 2x) ....Sine and cosine of complimentary angles are equal;
* 1 = sin π/2 ...............Trigo ratio of convenient angle: sin π/2 = sin 90o = 1
Thus, LHS = 2 sin 4x (cos 2x + 1)
= (2 sin 4x) [sin (π/2 - 2x) + sin π/2]
= (2 sin 4x){2 sin [(π/2 - 2x + π/2)/2] cos [(-2x)/2]}
= (4 sin 4x) sin (π/2 - x) cos (-x)
= (4 sin 4x) cos x cos x ..........[sin (π/2 - x) = cos x]
= 4 sin 4x cos2x
= RHS (Shown)
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If you ponder over Q1 above and its solutions, you should also be able to solve this:
Extra Q. Find the exact value of: sin 75o + sin 15o. (Answer: √(3/2))
P/s: If you don't know, email me and kindly introduce yourself when requesting something.
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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)
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3)
cos(A
+ B) ≡ cos A cos B – sin A sin B
(a)
Show that cos 2A ≡ 2 cos2 A – 1. (3)
(b)
Show that cos 4Ө ≡ 8 cos4 Ө – 8 cos2
Ө + 1. (3)
Hence
(c) solve, in radians to 3 significant figures, 10 cos4
Ө – 10 cos2 Ө + 1 = 0,
for 0 < Ө <
π/2, (6)
(d)
find…(Form 5: Integration) (5)
(28/1/2005 P2 Q10)
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4)
sin(A + B) ≡ sin A cos B + cos A sin
B
cos(A + B) ≡ cos A cos B – sin A sin B
(a)
Obtain an expression for cos 2Ө in terms of sin2
Ө. (2)
(b)
Obtain an expression for sin 2Ө in terms of sin Ө and
cos Ө. (1)
(c)
Show that cos 4Ө ≡ 1 – 8 sin2 Ө + 8 sin4
Ө. (4)
(d)
Solve, for 0 ≤ Ө ≤
π/2, the
equation sin2 Ө - sin4 Ө = 0.1, giving your solutions to
2 decimal places. (4)
(e)
Find…(Form 5: Integration) (5)
(27/5/2005 P2 Q10)
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5)
sin (A + B) ≡ sin A cos B + cos A sin B
cos (A + B) ≡ cos A cos B – sin A sin B
(f)
Obtain an expression for cos 2Ө in terms of cos2
Ө. (2)
(g)
Obtain an expression for sin 2Ө in terms of sin Ө and
cos Ө. (1)
(h)
Show that cos 3Ө ≡ 4 cos3 Ө - 3 cos Ө. (4)
(i)
Solve, for 0 ≤ Ө ≤ π/2, the equation 9 cos Ө – 12 cos3
Ө = 2, giving your answers to 3 significant figures. (4)
(j)
Find…(Form 5: Integration) (21/1/2008 P1 Q9)
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6)
Using cos(A + B) ≡ cos A cos B – sin A sin B,
(a)
show that
(i)
sin2 Ө = (1/2)
(ii)
cos2 Ө = (1/2)
f(Ө) = 1 + 10 sin2 Ө - 16 sin4 Ө .
(b)
Show that f(Ө) = 3 cos 2Ө – 2 cos 4Ө. (4)
(c)
Solve the equation
1 + 10 sin2 Өo - 16 sin4 Өo +
2 cos 4Өo = 0.25, for 0 ≤ Ө ≤ 180
giving your solutions to 1 decimal place. (4)
(d)
…(Form 5 ‘Integration’) (5)
(24/5/2006 P1 Q10)
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7)
cos(A + B) ≡ cos A cos B – sin A sin B,
f(Ө) = 5 cos Ө - 12 sin Ө
Given that f(Ө) = p cos (Ө + α), p > 0, 0 < α < π/2
(a)
(i) show that p
= 13
(ii) find, in
radians to 3 significant figures, the
value of α (5)
(b) Hence solve, to 2 significant figures, for 0
≤ Ө ≤ 2π, 5 cos Ө - 12 sin Ө = 9
(4)
(c) …(Integration …Form 5) (5)
(19/1/2007
P1 Q8)
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8)
cos(A + B) ≡ cos A cos B – sin A sin B
sin(A + B) ≡ sin A cos B + cos A sin B
(a) Write
down an expression for sin 2Ө in terms of sin Ө and cos Ө .
(1)
Show that:
(b)
sin2 Ө ≡ (1/2)(1 – cos 2Ө), (2)
(c)
sin2 (A + B) - sin2 (A - B) ≡ sin 2A sin 2B. (5)
Hence,
(d)
Show that (i)
sin2 3Ө - sin2 Ө ≡ sin 4Ө sin 2Ө
(ii) sin2 3Ө - sin2 Ө ≡
(1/2)(cos 2Ө – cos 6Ө). (4)
(e)
Find…(Form 5: Integration) (5)
(17/5/2007 P2 Q10)
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9)
Using the identities
cos(A + B) ≡ cos A cos B – sin A sin B
sin(A + B) ≡ sin A cos B + cos A sin B
express
(a) cos 2A in
terms of cos A. (2)
(b) sin 2A in
terms of sin A and cos A, simplifying your answer. (1)
(c) Hence
show that cos 3A ≡ 4 cos3 A – 3 cos A . (4)
(d) Solve,
for 0 ≤ x ≤ 180o, the equation 4 cos3 A – 3 cos A = 0.6,
giving your solutions to one decimal place. (5)
(e)
(Form 5: Integration) (5)
(12/5/2008 P1 Q9)
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10)
cos (A + B) ≡ cos
A cos B – sin A sin B
Show that
(a)
cos (A + B) + cos (A – B) ≡ 2 cos A cos B, (1)
(b)
cos 2A ≡ 2 cos2 A – 1 (2)
(c)
cos P + cos Q ≡ 2 cos [(P+Q)/2] cos [(P-Q)/2] (2)
(d)
Hence show that cos 8x + 2 cos 6x + cos 4x ≡ 4cos 6x
cos2 x (4)
(e)
(Form 5: Integration)… (6)
(17/5/2010 P2 Q9)
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11)
sin (A + B) ≡ sin A cos B + cos A sin B
cos (A + B) ≡ cos A cos B – sin A sin B
(a)
By writing tan (A + B) ≡ sin (A + B) / cos (A + B),
prove that tan (A + B) ≡ (tan A + tan B)/(1 – tan A tan B) (3)
(b)
Hence or otherwise, find the exact value of
(i)
tan 75o,
(ii)
tan 15o,
simplifying your answers as far as possible. (5)
(c)
Use the result in (a) to write down an expression for
tan 2Ө in terms of tan Ө.
(1)
(1)
(d)
Hence find the exact value of tan 22.5o. (4)
Given that tan Ө = 2/5 and Ө is an acute angle,
(e)
Find the exact value of sin 2Ө. (4)
(18/1/2010 P2 Q10)
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12)
cos(A
+ B) ≡ cos A cos B – sin A sin B
cos(A - B) ≡ cos A cos B + sin A sin B
(a)
Prove that cos 2A ≡ 2cos2 A – 1 (2)
f(Ө) = cos 5Ө + cos 3Ө + 2cos Ө
(b)
Show that
(i)
cos 5Ө + cos 3Ө ≡ 2cos 4Ө cos Ө,
(ii)
f(Ө) = 16cos5 Ө – 16cos3 Ө + 4cos
Ө. (6)
(c)
Hence or otherwise solve, for –π ≤ Ө ≤ π, giving the
value of Ө in terms of π,
the equation cos 5Ө + cos 3Ө - 2cos Ө = 0. (5)
the equation cos 5Ө + cos 3Ө - 2cos Ө = 0. (5)
(d)
(Form 5:
Intergration)… (5)
(14/5/2009 P2 Q10)
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13)
sin
(A + B) ≡ sin A cos B + cos A sin B
tan A ≡ sin A/cos A
(a)
Show that the equation
2 sin(x + α) = 5 sin(x – α)
Can be written in the form
3 tan x = 7 tan α (5)
(b)
Hence solve, to one decimal place, (5)
2 sin (2y + 50o) = 5 sin(2y -50o) for 0 ≤ y ≤ 180o
(17/01/2011 P2 Q7)
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14)
cos(A + B) ≡ cos A cos B – sin A sin B,
(a)
Show that
(i)
sin2 Ө =
(1 – cos 2Ө),
(ii)
cos2 Ө =
(cos 2Ө + 1) (3)
f(Ө) = 8 sin4 Ө + 4 sin2 Ө - 5.
(b)
Show that f(Ө) = cos 4Ө - 6 cos 2Ө. (4)
(c)
Solve, for 0 ≤ Ө ≤ π/2, the equation
4 sin4 Ө + 2 sin2 Ө + 3 cos 2Ө = 2.4
Give your solutions to 3 significant figures (4)
(d)
…(Form 5 ‘Integration’) (5)
(14/1/2011 P1 Q8)---------------------------------END---------------------------------------------------
2 comments:
Thank you for such useful questions. However,, many of them are very challenging for me.. If you can, could you please post the solutions for each question too?
Do you mind introducing yourself? Student? From where, please? I have already shared the solutions to all the questions with my offline students such as Kim SH, J. Min, etc. In due course, when time permits I will post them all in this blog. Thank you for your interest.
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