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Cos Θ - Analythic Trig - Q7.1234

by Zachary
(CA, USA)












































Given the following equation, use the inverse trigonometric function "arc cosine" (Cos-1) to compute a value for Θ.

Cos Θ = ½

Round the value of Θ to two decimal places, if appropriate.

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May 05, 2012
Analythic Trig Help - Q7.1234
by: Staff

Question:

by Zachary
(Campbell CA, USA)


Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)

Cos Θ = ½

list six specific solutions




Answer:


The diagram is very blurry, so I can’t read the drawing accurately.

Cos Θ = ½

Θ = cos⁻¹(1/2)

Θ = 60°, or 300°


>>> the “Principle Solutions” are:

Θ = 60° (1.05 radians), or 300° (5.24 radians)

60° = 1.05 radians
(60°/360°)*2pi = 1.0471975511966

300° = 5.24 radians
(300°/360°)*2pi = 5.235987755983






Thanks for writing.

Staff
www.solving-math-problems.com


Jul 08, 2012
6 specific solutions
by: Anonymous

Only 1/2 the problem was answered. The other half asked for 6 specific solutions.

Jul 08, 2012
Cos Θ - Analythic Trig - Q7.1234
by: Staff

Part II, 7/8/2012


Sorry, we originally overlooked the second part of the question. Thanks to the Anonymous blogger on 7/8/2012 for pointing that out.


Since no interval is given, we will use the Periodicity of the Cos function to find the General Solution using the integer k

k = any integer


θ = 1.05 ± 2πk, 5.24 ± 2πk

Calculation of Ten Specific Solutions (in radians)

2pi = 6.28


For k = 0

θ = 1.05 ± 2π*0, 5.24 ± 2π*0

θ = 1.05 ± 6.28*0, 5.24 ± 6.28*0

θ = 1.05 ± 0, 5.24 ± 0

θ = 1.05, 5.24




For k = 1

θ = 1.05 ± 2π*1, 5.24 ± 2π*1

θ = 1.05 ± 6.28*1, 5.24 ± 6.28*1

θ = 1.05 ± 6.28, 5.24 ± 6.28

θ = 7.33, -5.23, 11.52, -1.04


For k = 2

θ = 1.05 ± 2π*2, 5.24 ± 2π*2

θ = 1.05 ± 6.28*2, 5.24 ± 6.28*2

θ = 1.05 ± 12.56, 5.24 ± 12.56

θ = 13.61, -11.51, 17.8, -7.32



Ten Specific Solutions (in radians)

θ = 1.05, 5.24, 7.33, -5.23, 11.52, -1.04, 13.61, -11.51, 17.8, -7.32

 graph of Θ = cos⁻¹(1/2)







Thanks for writing.

Staff
www.solving-math-problems.com


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