Download Solving acosx + bsinx = c I. Solving sinx = C for 0 ≤ x

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Multilateration wikipedia , lookup

Line (geometry) wikipedia , lookup

System of polynomial equations wikipedia , lookup

History of trigonometry wikipedia , lookup

Trigonometric functions wikipedia , lookup

Transcript
Solving a cos x + b sin x = c
Dave L. Renfro
Central Michigan University
December 4, 2002
I. Solving sin x = C for 0
x<2
1. If jCj > 1, then sin x = C has no (real number) solutions.
[Note that the horizontal line whose equation is y = C does not intersect the unit circle.]
:
:
2. If sin x = 1, then x = 21 = 1:57. If sin x = 1, then x = 32 = 4:71.
[Note that the horizontal lines y = 1 and y = 1 each intersect the unit circle at one point..]
:
3. If sin x = 0, then x = 0;
= 0; 3:14.
[Note that the horizontal line y = 0 intersects the unit circle at two points.]
4. If sin x = C and 1 < C < 1, then you can obtain the reference angle for the solutions
using your calculator. Evaluate sin 1 C using you calculator in radian mode. Then the
reference angle will be the absolute value of this, sin 1 C . Use this reference angle
and what you know about the signs of SINE to determine the two values of x. (If
C > 0, then you’ll have a solution in quadrants I and II. If C < 0, then you’ll have a
solution in quadrants III and IV.)
[Note that the horizontal line y = C intersects the unit circle at two points.]
II. Solving cos x = C for 0
x<2
1. If jCj > 1, then cos x = C has no (real number) solutions.
[Note that the vertical line whose equation is x = C does not intersect the unit circle.]
:
2. If cos x = 1, then x = 0. If cos x = 1, then x = = 3:14.
[Note that the vertical lines x = 1 and x = 1 each intersect the unit circle at one point..]
:
3. If cos x = 0, then x = 21 ; 32 = 1:57; 4:71.
[Note that the vertical line x = 0 intersects the unit circle at two points.]
4. If cos x = C and 1 < C < 1, then you can obtain the reference angle for the solutions
using your calculator. Evaluate cos 1 C using you calculator in radian mode. Then
the reference angle will be than angle your calculator gives you (this will be when you
:
get a number between 0 and 12 = 1:57, which will happen when 0 < C < 1) or the
reference angle will be minus what your calculator gives you (this will be when you
:
:
get a number between 21 = 1:57 and = 3:14, which will happen when 1 < C < 0).
Use this reference angle and what you know about the signs of COSINE to determine
the two values of x. (If C > 0, then you’ll have a solution in quadrants I and IV. If
C < 0, then you’ll have a solution in quadrants II and III.)
[Note that the vertical line x = C intersects the unit circle at two points.]
III. Solving a cos x + b sin x = c for 0
x<2
1. Isolate one of the trigonometric functions on one side of the equation.
2. Square both sides of the equation.
3. Make use of the identity cos2 x+sin2 x = 1 to re–write the equation so that only SINE’s
appear or only COSINE’s appear.
4. Observe that you now have a quadratic equation relative to sin x or cos x. Solve the
quadratic equation aspect of the trigonometric equation. [If you get non–real roots, then
there’s not going to be a real solution to the original equation.]
5. At this point you will have either SINE equal to one or two numbers or COSINE equal
to one or two numbers. Solve for x according to the procedure on the other side. This
will give you the solution candidates.
6. Since we squared both sides at one point, we need to check for extraneous solutions.
To this end, plug each of the solution candidates into the original equation and discard
the candidates that don’t work.
IV. Example: Solve 3 cos x + 5 sin x = 5 for 0
1. 3 cos x = 5
2. 9 cos2 x = (5
sin2 x
3. 9 1
4. 16
5 sin x
5 sin x)2 = 25
= 25
50 sin x + 25 sin2 x
50 sin x + 25 sin2 x
50 sin x + 34 sin2 x = 0
)
x<2
sin x =
25
) 17 (sin x)2 25 (sin x) + 8 = 0
p
625
4 (17) (8)
25
9
8
=
= 1;
(2) (17)
34
17
:
5. sin x = 1 ) x = 12 = 1:57.
:
:
8
8
sin x = 17
) reference angle is sin 1 17
= 0:49. Hence, x = 0:49; 2:65.
:
Therefore, the solution candidates to the original equation are x = 0:49; 1:57; 2:65.
6. Plugging 0:49, 1:57, 2:65 into 3 cos x + 5 sin x gives, respectively, 5: 00013, 5: 0024,
:
0: 2846. Hence, the solutions to the original equation are x = 0:49; 1:57.
V. Some practice problems with answers
:
Answer: x = 3:96; 6:11.
:
2. 5 cos x + sin x = 2
Answer: x = 2:17; 4:51.
:
3. 5 cos x + 4 sin x = 6
Answer: x = 0:32; 1:03.
p
:
2 sin x = 5:73
4. 2 cos x
Answer: x = 3:99; 5:93.
1. 2 cos x
6 sin x = 3