cos 2x in .NET framework

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14.16 cos 2x
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cos x
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Substituting 2 cos x 1 for cos 2x, we have 2 cos2 x cos x cos x (2 cos x From cos x 0, x /2 and 3 /2; from cos x 1/2, x 2 /3 and 4 /3. The required solutions are x /2, 2 /3, 3 /2, and 4 /3.
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14.17 tan 2x
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sin 2x cos 2x 2 sin x cos x , we have cos 2x 2 sin x 2 sin x a cos x cos 2x 1b 2 sin x a 1 cos x cos 2x b cos 2x 1)(cos x 1) 0 0, x
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2 sin x cos x cos 2x
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From sin x 0, x 0, ; from cos x cos 2x cos x 2 cos2 x /3, 5 /3, and . The required solutions are x 0, /3, , and 5 /3.
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(2 cos x
14.18 sin 2x
cos 2x.
< 4 . Then 1, for which 2x /4, 5 /4, 9 /4, and /4, 5 /4, 9 /4, and
First Solution. Let 2x ; then we are to solve sin cos for 0 13 /4 and x /2 /8, 5 /8, 9 /8, and 13 /8 are the solutions. Second Solution. Dividing by cos 2x, the equation becomes tan 2x 13 /4 as in the first solution.
14.19 sin 2x
cos 4x.
cos 2(2x) 1 2 sin2 2x, the equation becomes sin 2x 1 (2 sin 2x 1)(sin 2x 1) 0 2 sin22x
Since cos 4x
CHAPTER 14 Trigonometric Equations
From 2 sin 2x 1 0 or sin 2x 1/2, 2x /6, 5 /6, 13 /6, and 17 /6 and x /12, 5 /12, 13 /12, and 17 /12; from sin 2x 1 0 or sin 2x 1, 2x 3 /2 and 7 /2 and x 3 /4 and 7 /4. All these values are solutions.
14.20 Solve the system
r sin r for r > 0 and 0 <2 . 4(1 sin )/3 or 4 sin2 3)(2 sin 4 sin 1) 0 3 and r 6. Note that 3 0 and 3 4(1 sin ) (1) (2)
Dividing (2) by (1), 1/sin
(2 sin
From 2 sin 1 0, sin 1/2 and 2 sin 3 0 is excluded since when r > 0, sin The required solutions are /6 and r
/6 and 5 /6; using (1), r(1/2) > 0 by (1). 5 /6 and r 6.
6 and
14.21 Solve Arccos 2x
Arcsin x.
If x is positive, Arccos 2x and Arcsin x terminate in quadrant I; if x is negative, terminates in quadrant II and terminates in quadrant IV. Thus, x must be positive. 21 x2. Taking the cosine of both members of the given equation, For x positive, sin x and cos b we have cos (Arccos 2x) Squaring, 4x2 1 x2, 5x2 1, and x cos (Arcsin x) 25/5 0.46 0.4472. cos or 2x 21 x2
Check: Arccos 2x radian.
Arccos 0.8944
Arcsin 0.4472, approximating the angle to the nearest hundredth
14.22 Solve Arccos (2x2
Let Arccos (2x2
2 Arccos 1. 2
1) and Arccos 1; then cos 2 2x2 1 and cos 2 A1 B 2
2 1 2 1 2.
Taking the cosine of both members of the given equation, cos Then 2x2
1 2.
cos 2
2 cos2
and x
1 2,
Check: For x
Arccos A
2 Arccos 1 or 2 /3 2
2( /3).
14.23 Solve Arccos 2x
Arccos x
If x is positive, 0 < Arccos 2x < Arccos x; if x is negative, Arccos 2x > Arccos x > 0. Thus, x must be negative. Let Arccos 2x and Arccos x; then cos 2x, sin a 21 4x2, cos b x, and sin b 21 x2 since both and terminate in quadrant II. Taking the cosine of both members of the given equation, cos ( or Squaring, 1 5x2 4x4
sin b 21
4x2 21 x2
x2 2x2
cos p>3
4x2 21 and x p/3.
1 2.
4x4, 3x2
3 4,
Check: Arccos ( 1)
Arccos A
2p/3
CHAPTER 14 Trigonometric Equations
1 4 p Arcsin
14.24 Solve Arcsin 2x
x. If x is negative, and terminate in
Let Arcsin 2x and Arcsin x; then sin 2x and sin quadrant IV; thus, x must be positive and acute. Taking the sine of both members of the given equation, sin a or Squaring, A 8 4 22 2x sin A 4p
bB x2
sin 4p cos b
1 2 22x
cos 4p sin b
1 2 22 21
A 2 22
1B x
1 B x2
x2, x2
1> A 10 0.26,
4 22 B and x
1 4p 1 4 (3.14)
0.2527. 0.79; and 0.53 0.79 0.26.
Check: Arcsin 0.5054
0.53; Arcsin 0.2527
14.25 Solve Arctan x
Arctan (1
Arctan 3.
x); then tan x and tan 1 x.
Arctan x and
Arctan (1
Taking the tangent of both members of the given equation, tan (a Then 3 4 4x
tan a tan b 1 tan a tan b 4x2, 4x2 Arctan A 1 4x
x 1 (2x
(1 x(1 1)2
x) x)
1 2.
tan A Arctan 4 B 3
0, and x 2(0.46)
Check: Arctan 2
2 Arctan 0.5000
0.92 and Arctan 3
Arctan 1.3333
SUPPLEMENTARY PROBLEMS
Solve each of the following equations for all x such that 0 ing approximate values for x.
14.26 sin x 23/2. 1/2. 0. 1) 1 0. 0. tan x 3. 3. 0. 0. 0. Ans. Ans. Ans. Ans. Ans. Ans. Ans. Ans. Ans. Ans. Ans. Ans. 2. 5 0. Ans. Ans.
x < 2 . Use Table 3 in Appendix 2 when find-
/3, 2 /3 /4, 3 /4, 5 /4, 7 /4 0, /2, , 3 /2 /4, 7 /6, 5 /4, 11 /6 /2, 7 /6, 11 /6 0, 2 /3, , 4 /3 /3, , 5 /3 0, /6, 5 /6, 0, /3, 5 /3 /6, /2, 5 /6 0, 3 /2 0 0, 1.96 , 4.22
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