Download Some harder examples of subsidiary angle form in trigonometry Yue

Some harder examples of subsidiary angle form in trigonometry
Yue Kwok Choy
1.
1 + sin θ
3 − cos θ
Find the maximum and minimum of
Solution Let
y=
1 + sin θ
.
3 − cos θ
3y – y cos θ = 1 + sin θ
sin θ + y cos θ = 3y – 1
1
y
3y − 1
sin θ +
cos θ =
2
2
1+ y
1+ y
1+ y2
1
cos α =
Let
1+ y
2
sin α =
,
sin (θ − α ) =
Since
y
1+ y2
sin θ cos α + cos θ sin α =
(1) becomes,
…. (1)
3y − 1
1+ y2
3y − 1
(☺ subsidiary angle form in L.H.S.)
1 + y2
sin (θ − α ) ≤ 1 , therefore
3y − 1
1 + y2
≤1
(3y − 1)2 ≤ 1 + y 2
9y2 – 6y + 1 ≤ 1 + y2
8y2 – 6y ≤ 0
y(4y – 3) ≤ 0
∴
0≤y≤
∴
Min. of
Note 1.
3
4
or
0≤
1 + sin θ 3
≤
.
3 − cos θ 4
1 + sin θ
= 0,
3 − cos θ
Max. of
If you are given to prove : 0 ≤
Solution
3
1 + sin θ
=
.
3 − cos θ
4
1 + sin θ 3
≤
3 − cos θ 4
1 + sin θ ≥ 0 and 3 – cos θ ≥ 3 – 1 > 0 ⇒ 0 ≤
Also, 4 sin θ + 3 cos θ = 5 sin (θ + α ) ≤ 5
∴ 4 + 4 sin θ ≤ 9 –3 cos θ
As
2.
1 + sin θ
3 − cos θ
, where
 3
α = tan −1   .
 4
⇒ 4(1 + sin θ) ≤ 3(3 – cos θ)
3 – cos θ > 0 , we have
1 + sin θ 3
≤
.
3 − cos θ 4
∴
0≤
1 + sin θ 3
≤
3 − cos θ 4
This is also a good question of application of differentiation in Calculus. ☺
1
2.
Solve :
8 cos x =
3
1
+
sin x cos x
Solution
Multiply the given equation by sin x cos x, we have:
8 sin x cos 2 x = 3 cos x + sin x
3
1
cos x + sin x
2
2
2 sin 2 x cos x =
sin 3x + sin x =
sin 3x =
1
3
sin x +
cos x
2
2
3
1
cos x − sin x
2
2
π
π
sin 3x = sin cos x − cos sin x
3
3
π

sin 3x = sin  − x 
3

∴
π

3x = 2nπ +  − x 
3

∴
x=
(☺ subsidiary angle form in R.H.S.)
1
π
nπ +
2
12
or
π

3x = 2nπ + π −  − x  , where n is an integer.
3

or
x = nπ +
π
3
, where n is an integer.
For the roots above , we check that sin x cos x ≠ 0 and they are roots of the given equation .
3.
Solve :
sin (π cos x ) = cos(π sin x )
Solution
sin (π cos x ) = cos(π sin x )
π

sin (π cos x ) = sin  − (π sin x )
2

∴
π

π cos x = 2nπ +  − (π sin x )
2

cos x ± sin x = 2n +
Now, since
∴
1
2
or
π

π cos x = 2nπ + π −  − (π sin x ) , where
2

n is an integer.
, where n is an integer.
π

cos x ± sin x = 2 cos x ±  ≤ 2
4

(☺ subsidiary angle form)
The only value for n is 0 .
π 1

2 cos x ±  =
4 2

The equation becomes :
x±
∴
or
π
2
= 2kπ ± cos −1
4
4
1
2

x =  2k ±  π ± cos −1
4
4

π
2

cos x ±  =
4
4

(use k here because n is used once)
,where
k is an integer.
2
4.
(a)
Given that
k is a constant, show that
π

π

y = sin  − x  + k cos − x 
2

3

can be expressed in the form y = r sin (x + α) , where
(b) What is the value of
α=
k when
2
π ?
3
r>0.
Plot its graph for this value of
α.
Solution
(a)
π
π
π

π



y = sin  − x  + k cos − x  = cos x + k  cos cos x + sin sin x 
3
3
2

3



=
=
3
 k
k sin x + 1 +  cos x
2
 2

3 2
k2 
3k
k+2
k + 1 + k + 
sin x +
cos x 
2
2
4
4  2 k + k +1
2 k + k +1

= k 2 + k + 1 sin (x + α )
where
(b) When
α=
(☺ subsidiary form)
k2 + k +1
r=
2
π,
3
tan
and
tan α =
k+2
3k
.
2
k+2
π=− 3=
.
3
3k
∴
–3k = k + 2,
k =−
1
2
2
Then
2 
3 
2 
 1  1

y =  −  +  −  + 1 sin  x + π  =
sin  x + π 
3  2
3 
 2  2


Graph :
y
2
1
−4
3π
−π
−2
3π
− π3
O
π
3
2π
3
π
4π
3
x
−1
−2
Exercise
1.
Find the relative maximum and relative minimum of
(Answer : relative maximum =
2.
Sketch the graph :
2 cos θ + 1
.
2 cos θ − 1
1
, relative minimum = 3 )
3
y = sin x + 3 cos x .
3