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Precalculus Mathematics Homework Help
Question of the basic properties of tan
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[QUOTE="thunderjolt, post: 4508766, member: 392741"] [h2]Homework Statement [/h2] I've solved a problem in my engineering homework to a point where I have the following data: ψ = arctan(3) = 71.57 (degrees) , inverse tangent of 3 δ = 30 degrees solve for gamma and chi (χ): tan(2gamma) = tan(2ψ)cos(δ) sin(2χ) = sin(2ψ)sin(δ) [h2]Homework Equations[/h2] tan(2gamma) = tan(2ψ)cos(δ) sin(2χ) = sin(2ψ)sin(δ) [h2]The Attempt at a Solution[/h2] My problem rests with the gamma solution. When I draw it (using other data in the problem), it is an angle larger than 70 degrees at least. When I solve for gamma with a calculator I get gamma = -16.5°. my process: tan(2ψ) = -.75 (-.75)cos(30) = -0.6495 arctan(-0.6495) = -33° gamma = -33/2 = -16.5° I know that there are properties of tangent that you need to take into consideration, but I just cannot remember them, and I am sure that is why I get this negative angle. Or, did I do this correctly, and I just need to redraw my problem? EDIT: the answer in the back of the book is gamma = 73.5 degrees [/QUOTE]
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Precalculus Mathematics Homework Help
Question of the basic properties of tan
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