Examples on Math ematical induction: Trigonometry Created by Mr. Francis Hung
1.
θ
(a)
Prove the identity: sin(k + 1)θ sin
(b)
Prove that sin x + sin 2x + … + sin nx =
(a)
LHS = sin(k + 1)θ sin
θ
2
+ sin
kθ
2
2
+ sin
sin
(k + 1)θ
kθ
sin = sin 2 2 sin 12 (n + 1)x sin 12 nx sin 12 x
Last updated: May 29, 2014
(k + 1)θ 2
sin
(k + 2)θ 2
.
for all positive integer n.
(k + 1)θ 2
1
3 1 1 1 θ = − cos k + θ − cos k + θ − cos k + θ − cos − 2 2 2 2 2 2 = − 1 cos k + 3 θ − cos 1 θ 2 2 2 = sin (b)
(k + 1)θ 2
sin
(k + 2 )θ 2
n = 1, LHS = sin x, RHS = ∴
=RHS
sin 12 (1 + 1)x sin 12 x sin 12 x
= sin x
LHS = RHS, it is true for n = 1.
Suppose sin x + sin 2x + … + sin kx =
sin 12 (k + 1)x sin 12 kx sin 12 x
Add sin (k+1)x to both sides, LHS = sin x + sin 2x + … + sin kx + sin (k+1)x sin 12 (k + 1)x sin 12 kx = sin (k+1)x + sin 12 x
( ) k + 1 x sin(12 )x + sin 12 k + 1 x sin 12 kx = sin sin 12 x =
sin( 12 k +) 1 x sin ( 12) k + 2 x sin 12 x
(by ( a))
= RHS If it is true for n = k then it is also true for n = k + 1. By the principle of mathematical induction, it is true for all positive integer n. ∴
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Page 1
Trigonometry 2.
Mr. Francis Hung
Prove that
1 2
n = 1, LHS =
RHS = =
+ cos x + cos 2x + … + cos nx = 1
sin (n + 12 )x 2 sin 12 x
for all positive integer n.
+ cos x
2
sin (1 + 12 )x 1 2
2 sin x
=
sin ( 32 )x 2 sin 12 x
3 sin x − 4 sin 3 12 x 1 2
2 sin 12 x 2 1
= 3 − 4 sin 2 x 2 3 − 2(1 − cos x ) = 2 1 + 2 cos x = = LHS 2 It is true for n = 1. sin (k + 12 )x 1 Suppose + cos x + cos 2x + … + cos kx = for some positive integer k. 2 2 sin 12 x When n = k + 1, 1 LHS = + cos x + cos 2x + … + cos kx + cos (k+1)x 2 sin (k + 12 )x = cos (k+1)x + 1
= = =
2 sin 2 x 2 cos(k + 1)x sin 12 x + sin (k + 12 )x 2 sin 12 x sin (k +
3 2
)x − sin ( )k − 12 (x +)sin
k + 12 x
2 sin 12 x sin (k + 1 + 2 sin 12 x
1 2
)x
= RHS
If it is true for n = k then it is also true for n = k + 1. By the principle of mathematical induction, it is true for all positive integer n. 1999 Paper 2 Q12 (a) Prove, by mathematical induction, that sin 2nθ cos θ + cos 3θ + cos 5θ + … + cos(2n – 1) θ = . 2 sin θ ∴
3.
where sin θ ≠ 0, for all positive integers n. (b)
Using (a) and the substitution θ =
π
− x , or otherwise, 2 sin 6 x show that sin x – sin 3x + sin 5x = , where cos x ≠ 0. 2 cos x
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Page 2
Trigonometry 4.
Mr. Francis Hung
Prove that
1
tan
2
x
Note that tan 2θ = 1
n = 1, LHS =
2 x
+
2
1 22
tan
2 tan θ 1 − tan 2 θ
tan
x
+L+
22 →
1 2n
cot x =
tan
x
2n
1 − tan 2
= x 2
2 tan x
2
1 2n
cot
x
2n
1 x = cot 2 2
− cot x
−
1 2
tan
where x≠mπ for n = 1,2, … x
2
...............(*)
x
2
1 RHS = cot − cot x 2 2 = LHS by (*) It is true for n = 1. 1 1 1 1 x x x x Suppose tan + tan +L+ tan = cot − cot x 2 2 22 22 2k 2k 2k 2k 1 1 1 1 x x x x When n = k + 1, LHS = tan + tan +L+ tan + tan 2 2 22 22 2k 2 k 2 k +1 2 k +1 1 1 x x = k tan k + k cot k − cot x 2 +1 2 +1 2 2 1 1 x 1 x = ⋅ tan + k cot k − cot x k k +1 2 2 2 2 2 1 1 x x 1 x = ⋅ cot − cot + cot k − cot x , by (*) 2k 2 2 k +1 2k 2k 2 1 x = cot − cot x = RHS k +1
k +1
2 It is also true for n = k + 1
2
By the principle of mathematical induction, the formula is true for all positive integer n. n
5.
Prove that
∑ cot (2r ) = tan −1
2
−1
(2n + 1) −
r =1
1
n
π=
4
∑ tan r =1
−1
1 2 for all positive integer n. 2r
1 cot θ = ⇒ cot x = tan tan θ x n n 1 −1 −1 2 ∴ ∑ cot (2r ) = ∑ tan 2 2r r =1 r =1 1
–1
tan −1 (2n + 1) −
1
π = tan
–1
−1
(2n + 1) − tan 1 1 = tan −
−1
2n + 1 − 1
= tan −1
n
n +1 1 n Let P(n) ≡ “ tan −1 2 = tan −1 .” for all positive integers n. n +1 2r r =1
4
1 + 2n + 1
n
∑
1 –1 1 2 = tan = RHS 2r 2 k 1 k Suppose ∑ tan −1 2 = tan −1 k +1 2r r =1 k +1 1 1 k 1 1 k –1 −1 − = tan −1 + tan–1 n = k + 1, ∑ tan 2 = ∑ tan 1 2 + tan 2 2 k +1 2(k + 1) 2(k + 1) 2r r =1 2r r =1 2 kk+1 + 2(k1+1) = tan −1 (k + 1)(2k + 2k + 1) = tan −1 k + 1 = tan −1 1 − (kk+1) ⋅ 1 2 k 3 + 6 k 2 + 5k + 2 k+2 2 ( k +1) 1
n = 1,
∑ tan
−1
r =1
2
2
It is also true for n = k + 1. By MI, the formula is true for all positive integer n. http://www.hkedcity.net/ihouse/fh7878/
Page 3
Trigonometry
Mr. Francis Hung n
6.
Prove that
6.
n = 1,
dy
d sin x dx n =
dx
= sin x +
cos x = sin x +
nπ for all positive integer n. 2
π
, it is true for n = 1
2
k
kπ for some positive integer k. 2 dx k d +1 y kπ (k + 1)π = cos x + , it is also true for n = k + 1 = sin x + k +1 2 2 dx
Suppose
d y k
= sin x +
n
By the principle of mathematical induction, d ny dx 7.
Prove that cos 2 x cos 2 2 x L cos 2 n x = n = 1, LHS = cos 2x, RHS =
=
sin 2 n +1 x 2 n sin 2 x
= sin x+
nπ for n = 1, 2, 3, … 2
, where sin 2x ≠ 0 for all positive integer n.
sin 4 x 2 sin 2 x 2 sin 2 x cos 2 x 2 sin 2 x
= cos 2x = LHS It is true for n = 1 Suppose it is true for n = k, where k is a positive integer. 2
i.e. cos 2x cos 2 x
sin 2 k +1 x
k
……
cos 2 x = 2 k sin 2 x ...........(*) 2
k
k+1
When n = k + 1, LHS = cos 2x cos 2 x …… cos 2 x cos 2 = = =
sin 2
k +1
x
2 k sin 2 x
cos 2
k+1
x
x by (*)
2 sin 2 k +1 x cos 2 k +1 x 2 k +1 sin 2 x sin 2 k + 2 x 2 k +1 sin 2 x
= RHS
It is also true for n = k + 1 By the principle of mathematical induction, the formula is true for all positive integer n.
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Page 4
Trigonometry
8.
(a)
Mr. Francis Hung
Prove that sin(θ +
k −1
2
α)
Hence prove that sin(θ + (b)
sin
k −1
2
kα
2 α)
+ sin sin
α
2
kα
2
sin(θ + kα) =
+ sin
(b)
2
2
[cos(θ –
α
2
) – cos(θ +
sin(θ + kα) = sin(θ +
kα
2
)sin
2k + 1 2
k +1
α)]
α
2
Prove by mathematical induction that sin θ + sin(θ + α) + …… + sin[θ+(n – 1)α] =
(a)
α
1
sin θ + ( n2−1 )α sin n2α sin α2
sin(θ + k − 1 α) sin kα + sin α sin(θ + kα) 2 2 2 1 2k − 1 α 1 2k + 1 2k − 1 = – [cos(θ + α) – cos(θ – )] – [cos(θ + α) – cos(θ + α)] 2 2 2 2 2 2 1 2k + 1 α = [cos(θ – ) – cos(θ + α)] 2 2 2 α k −1 kα sin(θ + + sin sin(θ + kα) α) sin 2 2 2 1 2k + 1 α = – [cos(θ + α) – cos(θ – )] 2 2 2 1 kα k +1 = – (–2)[sin(θ + ) sin(θ + α)] 2 2 2 k +1 kα = sin(θ + )sin α 2 2 sin θ sin α2 = sin θ n = 1, LHS = sin θ, RHS = sin α2 It is true for n = 1. Suppose it is true for n = k. i.e. sin θ + sin(θ + α) + …… + sin[θ+(k – 1)α] =
sin θ + ( k2−1 )α sin k2α sin α2
...........(*)
When n = k + 1, LHS = sin θ + sin(θ + α) + …… + sin[θ+(k – 1)α] + sin (θ+ k α) sin θ + ( k2−1 )α sin k2α = + sin (θ+ k α) by (*) sin α2 = =
sin θ + ( k2−1 )α sin k2α + sin α2 sin (θ + kα ) sin α sin (θ +
kα 2
)sin k 2 1 α
sin α2
+
2
by (a)
= RHS Hence it is also true for n = k + 1 By the principle of induction, the statement is true for all positive integer n.
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Page 5
Trigonometry
9.
(a)
Mr. Francis Hung
Prove that cos
(k + 1)π 2n − 1
Hence show that cos (b)
sin
kπ
2n − 1
(k + 1)π 2n − 1
sin
+ cos kπ
2n − 1
2(k + 1)π
π
sin
2n − 1
2n − 1
2(k + 1)π
+cos
2n − 1
1
=
2
[sin
π
sin
2n − 1
(2k + 3)π
−
2n − 1
= cos
(k + 2)π 2n − 1
sin sin
π
2n − 1
]
(k + 1)π 2n − 1
Prove by mathematical induction on m that
cos
2π 2n − 1
+ cos
4π 2n − 1
+ … + cos
cos
2mπ
(m + 1)π 2n − 1
=
2n − 1
mπ
sin
2n − 1 .
π
sin
2n − 1 (c)
Hence find the value of cos
(a)
cos = =
1 2 1 2
(k + 1)π 2n − 1
[sin [sin
cos
kπ
sin
(2k + 1)π
2n − 1 −
2n − 1
(2k + 3)π
−
2n − 1
(k + 1)π 2n − 1
sin
sin sin
2π
+ cos
2n − 1
2n − 1
2(k + 1)π 2n − 1 + sin
2n − 1
k
+ cos
2n − 1
4π 2n − 1
sin
+ … + cos
2(n − 1)π 2n − 1
.
π
2n − 1
(2k + 3)π
−
2n − 1
sin
(2k + 1)π 2n − 1
]
]
+ cos
2(k + 1)π 2n − 1
sin
π
2n − 1
= 1 [sin (2k + 3)π − sin ] = cos (k + 2 )π sin (k + 1) 2 2n − 1 2n − 1 2n − 1 2n − 1
(b)
m = 1, LHS = cos
cos
2π 2n − 1
, RHS =
2π
π
sin
2n − 1 = cos 2π π 2n − 1 sin 2n − 1
2n − 1
It is true for m = 1. Suppose it is true for m = k, where k is a positive integer.
i.e. cos
2π 2n − 1
+ cos
4π 2n − 1
+ … + cos
2kπ 2n − 1
cos =
(k + 1)π 2n − 1 sin
sin π
kπ
2n − 1 .
2n − 1
When m = k + 1, LHS = cos
cos =
2π 2n − 1
(k + 1)π
+ cos
sin
4π 2n − 1
+ … + cos
2kπ 2n − 1
+ cos
2(k + 1)π 2n − 1
kπ
2n − 1 + cos 2(k + 1)π by induction assumption π 2n − 1 sin 2n − 1
2n − 1
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Page 6
Trigonometry
Mr. Francis Hung
cos
(k) + 1 π 2n − 1
=
kπ ( )
sin
2n − 1 sin
cos
(k) + 2 π( ) 2n − 1
=
sin
sin
+ cos
2 k +1 π 2n − 1
sin
π
2n − 1
π
2n − 1
k +1 π
2n − 1
π
by (a)
2n − 1
= RHS Hence it is also true for m = k + 1 By the principle of mathematical induction, it is true for all positive integer m. (c)
Put m = n – 1, cos
cos =
2π 2n − 1 nπ
2n − 1 sin
sin =
+ cos
sin
sin π − sin =
= 2 sin
2(n − 1)π 2n − 1
(n − 1)π 2n − 1
2n − 1
2n − 1 2 sin
0 − sin
+ … + cos
π
(2n − 1)π
2 sin
4π 2n − 1
− sin π
π
2n − 1
2n − 1 π
2n − 1 π
2n − 1 π
2n − 1 π
2n − 1
1 =− 2
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Page 7
Trigonometry 10.
(a)
Mr. Francis Hung
Prove by mathematical induction that for any positive integer n,
(− 1)n 1 sin 2nθ +
sin θ – sin 3θ + sin 5θ – ..... + (–1) n+1 sin(2n – 1)θ = (b)
2 cos θ
, where cos θ ≠ 0.
Using (a), or otherwise, find the general solution of sin θ – sin 3θ + sin 5θ – ..... + (–1) n+1 sin(2n – 1)θ = 0
11.
Given sin α (a)
≠
0, and n is a positive integer.
Express the product of
(b )
(c )
⋅ cos 2 k
sin
functions in terms of k and
α
as a sum or difference of two trigonometric
.
Without using mathematical induction, show that (i )
sin α ⋅ cos 2α + sin α cos 4 α + sin α cos 6α + ⋅ ⋅ ⋅ + sin α cos 2 nα =
(ii)
cos 2 α
+
cos 4 α
+
cos 6 α
+ ⋅⋅⋅+
cos 2 n α
=
cos (n
1 2
+ 1 )α ⋅ sin
sin
[sin (2n + 1)α − sin α ] ; nα
α
.
Hence or otherwise, prove that
sin 2
α+
sin 2 2α + sin 2 3α + ⋅ ⋅ ⋅ + sin 2 nα =
(Given that sin 2 A =
1 − cos 2 A
(a)
sin α ⋅ cos 2kα =
(b)
(i)
2 1
n
2
−
sin nα ⋅ cos(n + 1)α 2 sin α
)
[sin(2k + 1)α − sin(2k − 1)α ]
2 sin α cos 2α + sin α cos 4α + sin α cos 6α + … + sin α cos 2nα 1
1
1
1
2
2
2
2
= (sin 3α − sin α ) + (sin 5α − sin 3α ) + (sin 7α − sin 5α ) + ⋅ ⋅ ⋅ + [sin(2n + 1)α − sin α ] = (ii)
2
[sin(2n + 1)α − sin α ]
cos 2α + cos 4α + cos 6α + … + cos 2nα= =
(c)
1
sin(2n + 1)α − sin α 2 sin α
cos(n + 1)α ⋅ sin nα
sin α sin 2 α + sin 2 2α + sin 2 3α + L + sin 2 2nα 1 − cos 2α 1 − cos 4α 1 − cos 6α 1 − cos 2nα = + + +…+ 2 2 2 2 n cos 2α + cos 4α + cos 6α + L + cos 2nα =2 − 2 =
n
2
−
cos(n + 1)α ⋅ sin nα 2 sin α
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Page 8
Trigonometry 25.
Mr. Francis Hung
Prove , by mathematical induction, that for all positive integers n, sin (n + 12 )x − sin 12 x cos x + cos 2x + cos 3x + … + cos nx = where sin 12 x ≠ 0. 2 sin 12 x n = 1, LHS = cos x sin (1 + 12 )x − sin 12 x sin ( 32 )x − sin 12 x RHS = = 2 sin 12 x 2 sin 12 x =
3 sin 12 x − 4 sin 3 12 x − sin 12 x 2 sin 12 x 3 − 4 sin
= = =
2 1
x −1
2
2 2 − 2(1 − cos x ) 2 2 cos x
= LHS 2 It is true for n = 1. Suppose
cos x + cos 2x + … + cos kx =
sin (k +
1 2
)x − sin 12 x
2 sin 12 x
for some positive integer k.
When n = k + 1, LHS = cos x + cos 2x + … + cos kx + cos (k+1)x sin (k + 12 )x − sin 12 x = + cos (k+1)x 2 sin 12 x = = =
sin k +
1
(
2
x − sin 1 x + 2 cos k + 1 x sin 1 x
)
2
2 sin 12 x
sin (k + 12 )x + sin (k +
3 2
(
)
2
)x − sin (k + 12 )x − sin 12 x
2 sin 12 x sin (k + 1 + 12 )x − sin 12 x 2 sin 12 x
= RHS
If it is true for n = k then it is also true for n = k + 1. By the principle of mathematical induction, it is true for all positive integer n. ∴
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Page 9
Trigonometry 26.
Mr. Francis Hung
Prove , by mathematical induction, that for all positive integers n, sin (n + 1)x − sin nx − sin x sin x + sin 2x + sin 3x + … + sin nx = where cos x ≠ 1. 2(cos x − 1) n = 1, LHS = sin x sin (1 + 1)x − sin x − sin x RHS = 2(cos x − 1) 2 sin x cos x − 2 sin x = 2(cos x − 1) 2 sin x(cos x − 1) = 2(cos x − 1) = sin x = LHS Suppose it is true for some positive integer k. i.e. sin (k + 1)x − sin kx − sin x sin x + sin 2x + sin 3x + … + sin kx = for some positive integer k. 2(cos x − 1) When n = k + 1, LHS = sin x + sin 2x + sin 3x + … + sin kx + sin (k + 1)x sin (k + 1)x − sin kx − sin x = + sin (k + 1)x (by induction assumption) 2(cos x − 1) (sin ) k + 1 x − sin kx(− sin)(x )+ 2 cos x − 1 sin k + 1 x = 2(cos x − 1) ()sin k + 1 x − sin kx − sin (x) + 2(cos ) x sin k + 1 x − 2 sin k + 1 x = 2(cos x − 1)
) k + 1 x (cos) x − sin k + 1 x − sin x = − sin kx + 2( sin 2(cos x − 1) ) k + 2 x +(sin ) kx − sin k + 1 x − sin x − sin kx +( sin = 2(cos x − 1) sin ( ) k + 2 (x −)sin k + 1 x − sin x = = RHS 2(cos x − 1) If it is true for n = k then it is also true for n = k + 1. By the principle of mathematical induction, it is true for all positive integer n. ∴
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Page 10
Trigonometry 27.
Mr. Francis Hung
Prove , by mathematical induction, that for all positive integers n, 1 − cos 2nx sin x + sin 3x + sin 5x + … + sin (2n – 1)x = where sin x ≠ 0. 2 sin x n = 1, LHS = sin x 1 − cos 2 x RHS = 2 sin x 1 − (1 − 2 sin 2 x ) = 2 sin x = sin x = LHS It is true for = 1.for some positive integer k. i.e. Suppose it isntrue 1 − cos 2 kx sin x + sin 3x + sin 5x + … + sin (2k – 1)x = 2 sin x When n = k + 1, LHS = sin x + sin 3x + sin 5x + … + sin (2k – 1)x + sin (2k + 1)x 1 − cos 2kx = + sin (2k + 1)x (induction assumption) 2 sin x 1 − cos 2kx + 2 sin (2k + 1)x sin x = 2 sin x 1 − cos 2kx − [cos(2k + 2 )x − cos 2kx ] = 2 sin x 1 − cos(2k + 2 )x = = RHS 2 sin x If it is true for n = k then it is also true for n = k + 1. By the principle of mathematical induction, it is true for all positive integer n. ∴
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Page 11
