# 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.

### Comments for Cos Θ - Analythic Trig - Q7.1234

 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 = ½list six specific solutions Answer:The diagram is very blurry, so I can’t read the drawing accurately.Cos Θ = ½Θ = 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 = 1.0471975511966300° = 5.24 radians(300°/360°)*2pi = 5.235987755983Thanks 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, 7/8/2012Sorry, 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 integerθ = 1.05 ± 2πk, 5.24 ± 2πkCalculation of Ten Specific Solutions (in radians)2pi = 6.28For k = 0θ = 1.05 ± 2π*0, 5.24 ± 2π*0θ = 1.05 ± 6.28*0, 5.24 ± 6.28*0θ = 1.05 ± 0, 5.24 ± 0θ = 1.05, 5.24For k = 1θ = 1.05 ± 2π*1, 5.24 ± 2π*1θ = 1.05 ± 6.28*1, 5.24 ± 6.28*1θ = 1.05 ± 6.28, 5.24 ± 6.28θ = 7.33, -5.23, 11.52, -1.04For k = 2θ = 1.05 ± 2π*2, 5.24 ± 2π*2θ = 1.05 ± 6.28*2, 5.24 ± 6.28*2θ = 1.05 ± 12.56, 5.24 ± 12.56θ = 13.61, -11.51, 17.8, -7.32Ten 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 Thanks for writing. Staff www.solving-math-problems.com