Cos 10 Cos 30 Cos 50 Cos 70 2b211
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 2z6p3t
Overview 5o1f4z
& View Cos 10 Cos 30 Cos 50 Cos 70 as PDF for free.
More details 6z3438
- Words: 993
- Pages: 2
cos 10° cos 30° cos 50° cos 70° Yue Kwok Choy
1.
Product sum formula Analysis : (a)
cos 30° is connected with special angle and should be singled out.
(b)
You can then combine any two of the three
:
cos 10° , cos 50°, cos 70°
to form a product.
Then change this to a sum. Steps : cos 10° cos 30° cos 50° cos 70° = cos 30° cos 70° (cos 50° cos 10°)
2.
(
)
=
3 1 3 ⎛1 ⎞ cos 70o × cos 60o + cos 40o = cos 70o × ⎜ + cos 40o ⎟ 2 2 4 ⎝2 ⎠
=
3 ⎡1 3 ⎡1 1 ⎤ ⎤ cos 70o + cos 70o cos 40o ⎥ = cos 70o + cos 110o + cos 30o ⎥ ⎢ ⎢ 4 ⎣2 2 ⎦ ⎦ 4 ⎣2
=
3 ⎡1 1⎛ 3 ⎞⎟⎤ 3 ⎡ 3⎤ 3 o o = ⎢ cos 70 + ⎜⎜ − cos 70 + ⎥ ⎢ ⎥= 4 ⎢⎣ 2 2⎝ 2 ⎟⎠⎥⎦ 4 ⎢⎣ 4 ⎥⎦ 16
(
)
Vieta’s theorem Analysis : (a)
We should think ‘backward’. So
(b)
cos 3A =
Let A = 10°.
It is not ‘special’. However, 3A = 30°
3 and we may think of solving the equation : 2
is special.
2 cos 3A − 3 = 0
The given is a product and we therefore need the Vieta’s theorem, which can relate the product of roots with the coefficients of the equation. Note that in the calculation below, we get only a cubic equation.
Steps : (a)
⇒ (b)
(c)
⇒
2 cos 3A − 3 = 0
cos 3A =
3 2
3A = 30°, 330° , 390° (other roots neglected)
2 cos 3A − 3 = 0
⇒
(
)
⇒
8 cos 3 A − 6 cos A − 3 = 0
⇒
x = cos A is a root of
(
)(
A = 10°, 110° , 130°
8x 3 − 6x − 3 = 0
x = cos 10°, cos 110° cos 130° are roots of By Vieta’s Theorem,
⇒
2 4 cos A − 3 cos A − 3 = 0 3
8x 3 − 6x − 3 = 0 .
cos 10o cos 110o cos 130o = −
)
cos 10 o − cos 70 o − cos 50 o =
3 8
∴
cos 10o cos 50 o cos 70 o =
3 8
∴
cos 10 o cos 30 o cos 50 o cos 70 o =
coeff. of constant term 3 = coeff. of x 3 term 8
3 3 3 = 8 2 16 1
3.
Complex number
Analysis : (a)
Naturally, we like to begin with z = cos 10° + i sin 10° 3
z = cos 30° + i sin 30° , z = cos 50° + i sin 50° , z7 = cos 70° + i sin 70°.
theorem, since (b)
and obviously we have to use de Moivres’
5
During the calculation in part (b) below, we find that it is rather difficult to simplify. We therefore need more relations to ‘lower’ the power. Equations
(1) and (2) are then found.
Steps : (a)
⇒
z = cos 10° + i sin 10°
1/z = cos 10° – i sin 10°
1⎛ 1 ⎞ z2 + 1 cos 10o = ⎜ z + ⎟ = , 2⎝ z⎠ 2z cos 50o =
1⎛ 1 ⎞ z6 + 1 cos 30 o = ⎜ z 3 + 3 ⎟ = 2⎝ z ⎠ 2z 3
1 ⎛ 5 1 ⎞ z10 + 1 , ⎜z + 5 ⎟ = 2⎝ z ⎠ 2z 5
(b)
cos 10 o cos 30 o cos 50 o cos 70 o =
(c)
It is difficult to simplify (b).
z 3 = cos 30o + i sin 30 o = z 6 = cos 60 o + i sin 60o =
(d)
1 ⎛ 7 1 ⎞ z14 + 1 ⎜z + 7 ⎟ = 2⎝ z ⎠ 2z 7
(
)(
)(
)(
)
z 2 + 1 z 6 + 1 z10 + 1 z14 + 1 z 2 + 1 z 6 + 1 z10 + 1 z14 + 1 × × × = 2z 2z 3 2z 5 2z 7 16z16
But note that : ⇒
z = cos 10° + i sin 10°
cos 70o =
z9 = cos 90° + i sin 90°
⇒
z9 = i
....
(1)
⇒
z6 = iz3 + 1
....
(2)
3 1 + i, 2 2 ⎛ 3 1 ⎞ 1 3 + i = i⎜ + i⎟ + 1 ⎜ 2 2 2 2 ⎟⎠ ⎝
Substitute (1) in part (b), cos 10 o cos 30 o cos 50 o cos 70 o = =
(z
2
=
(z
2
(z
6
)(
)(
)(
)(
) [(
)(
(z
2
)(
)
+ 1 z 6 + 1 − iz 3 − 1 + iz 5 + iz + 1 16iz 7
) (
)
(
) (
)(
)(
)
+ 1 z 6 + 1 (iz + 1) iz 5 + 1 z 2 + 1 z 6 + 1 − z 6 + iz 5 + iz + 1 = 7 16iz 16iz 7
, by (2)
)(
)(
)
+ 1 z 6 + 1 − iz 3 + iz 5 + iz z2 + 1 z6 + 1 − z2 + z4 + 1 = 7 16iz 16z 6
)]
+ 1 z2 + 1 z4 − z2 + 1
=
16z 6
(z =
6
(
)(
)
+ 1 z6 + 1 1 ⎛ z6 + 1⎞ ⎜ ⎟ = 16z 6 4 ⎜⎝ 2z 3 ⎟⎠
1 = cos 30o 4
)
2
2
2
1 ⎛ 3 ⎞⎟ 3 = ⎜ = 4 ⎜⎝ 2 ⎟⎠ 16
2
1.
Product sum formula Analysis : (a)
cos 30° is connected with special angle and should be singled out.
(b)
You can then combine any two of the three
:
cos 10° , cos 50°, cos 70°
to form a product.
Then change this to a sum. Steps : cos 10° cos 30° cos 50° cos 70° = cos 30° cos 70° (cos 50° cos 10°)
2.
(
)
=
3 1 3 ⎛1 ⎞ cos 70o × cos 60o + cos 40o = cos 70o × ⎜ + cos 40o ⎟ 2 2 4 ⎝2 ⎠
=
3 ⎡1 3 ⎡1 1 ⎤ ⎤ cos 70o + cos 70o cos 40o ⎥ = cos 70o + cos 110o + cos 30o ⎥ ⎢ ⎢ 4 ⎣2 2 ⎦ ⎦ 4 ⎣2
=
3 ⎡1 1⎛ 3 ⎞⎟⎤ 3 ⎡ 3⎤ 3 o o = ⎢ cos 70 + ⎜⎜ − cos 70 + ⎥ ⎢ ⎥= 4 ⎢⎣ 2 2⎝ 2 ⎟⎠⎥⎦ 4 ⎢⎣ 4 ⎥⎦ 16
(
)
Vieta’s theorem Analysis : (a)
We should think ‘backward’. So
(b)
cos 3A =
Let A = 10°.
It is not ‘special’. However, 3A = 30°
3 and we may think of solving the equation : 2
is special.
2 cos 3A − 3 = 0
The given is a product and we therefore need the Vieta’s theorem, which can relate the product of roots with the coefficients of the equation. Note that in the calculation below, we get only a cubic equation.
Steps : (a)
⇒ (b)
(c)
⇒
2 cos 3A − 3 = 0
cos 3A =
3 2
3A = 30°, 330° , 390° (other roots neglected)
2 cos 3A − 3 = 0
⇒
(
)
⇒
8 cos 3 A − 6 cos A − 3 = 0
⇒
x = cos A is a root of
(
)(
A = 10°, 110° , 130°
8x 3 − 6x − 3 = 0
x = cos 10°, cos 110° cos 130° are roots of By Vieta’s Theorem,
⇒
2 4 cos A − 3 cos A − 3 = 0 3
8x 3 − 6x − 3 = 0 .
cos 10o cos 110o cos 130o = −
)
cos 10 o − cos 70 o − cos 50 o =
3 8
∴
cos 10o cos 50 o cos 70 o =
3 8
∴
cos 10 o cos 30 o cos 50 o cos 70 o =
coeff. of constant term 3 = coeff. of x 3 term 8
3 3 3 = 8 2 16 1
3.
Complex number
Analysis : (a)
Naturally, we like to begin with z = cos 10° + i sin 10° 3
z = cos 30° + i sin 30° , z = cos 50° + i sin 50° , z7 = cos 70° + i sin 70°.
theorem, since (b)
and obviously we have to use de Moivres’
5
During the calculation in part (b) below, we find that it is rather difficult to simplify. We therefore need more relations to ‘lower’ the power. Equations
(1) and (2) are then found.
Steps : (a)
⇒
z = cos 10° + i sin 10°
1/z = cos 10° – i sin 10°
1⎛ 1 ⎞ z2 + 1 cos 10o = ⎜ z + ⎟ = , 2⎝ z⎠ 2z cos 50o =
1⎛ 1 ⎞ z6 + 1 cos 30 o = ⎜ z 3 + 3 ⎟ = 2⎝ z ⎠ 2z 3
1 ⎛ 5 1 ⎞ z10 + 1 , ⎜z + 5 ⎟ = 2⎝ z ⎠ 2z 5
(b)
cos 10 o cos 30 o cos 50 o cos 70 o =
(c)
It is difficult to simplify (b).
z 3 = cos 30o + i sin 30 o = z 6 = cos 60 o + i sin 60o =
(d)
1 ⎛ 7 1 ⎞ z14 + 1 ⎜z + 7 ⎟ = 2⎝ z ⎠ 2z 7
(
)(
)(
)(
)
z 2 + 1 z 6 + 1 z10 + 1 z14 + 1 z 2 + 1 z 6 + 1 z10 + 1 z14 + 1 × × × = 2z 2z 3 2z 5 2z 7 16z16
But note that : ⇒
z = cos 10° + i sin 10°
cos 70o =
z9 = cos 90° + i sin 90°
⇒
z9 = i
....
(1)
⇒
z6 = iz3 + 1
....
(2)
3 1 + i, 2 2 ⎛ 3 1 ⎞ 1 3 + i = i⎜ + i⎟ + 1 ⎜ 2 2 2 2 ⎟⎠ ⎝
Substitute (1) in part (b), cos 10 o cos 30 o cos 50 o cos 70 o = =
(z
2
=
(z
2
(z
6
)(
)(
)(
)(
) [(
)(
(z
2
)(
)
+ 1 z 6 + 1 − iz 3 − 1 + iz 5 + iz + 1 16iz 7
) (
)
(
) (
)(
)(
)
+ 1 z 6 + 1 (iz + 1) iz 5 + 1 z 2 + 1 z 6 + 1 − z 6 + iz 5 + iz + 1 = 7 16iz 16iz 7
, by (2)
)(
)(
)
+ 1 z 6 + 1 − iz 3 + iz 5 + iz z2 + 1 z6 + 1 − z2 + z4 + 1 = 7 16iz 16z 6
)]
+ 1 z2 + 1 z4 − z2 + 1
=
16z 6
(z =
6
(
)(
)
+ 1 z6 + 1 1 ⎛ z6 + 1⎞ ⎜ ⎟ = 16z 6 4 ⎜⎝ 2z 3 ⎟⎠
1 = cos 30o 4
)
2
2
2
1 ⎛ 3 ⎞⎟ 3 = ⎜ = 4 ⎜⎝ 2 ⎟⎠ 16
2
Related Documents c2h70
Cos 10 Cos 30 Cos 50 Cos 70 2b211
July 2021 0
Cos Me Ti Cos 465h68
September 2022 0
Cos Handouts 5l4i2o
February 2022 0
Cos Mart 1toa
December 2021 0
Cos To 2p6153
January 2023 0
Chico Cos 3e3kf
July 2021 0More Documents from "DavidMotta" 1p6663