Really struggling with trig functions, can anyone give me a step by step how to find angles between 0 and 360 please :)

Thanks,

Kris

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- April 1st 2013, 05:19 AMKrislton2 cos \theta - 4 sin^2 \theta = 0
Really struggling with trig functions, can anyone give me a step by step how to find angles between 0 and 360 please :)

Thanks,

Kris - April 1st 2013, 05:46 AMProve ItRe: 2 cos \theta - 4 sin^2 \theta = 0
- April 1st 2013, 05:54 AMHallsofIvyRe: 2 cos \theta - 4 sin^2 \theta = 0
First, you need to know some trig identities- like sin^2x+ cos^2x= 1. In this particular problem, that lets you replace " " with [tex]1- cos^2(\theta)[/itex] so the problem becomes [tex]2cos(\theta)- 4(1- cos^2(\theta))= 4cos^2(\theta)+ 2 cos(\theta)- 4= 0[/itex].

Do you recognize that as a quadratic equation? If you replace " " with "y" you have [tex]4y^2+ 2y- 4= 0[/quote]. You can use the quadratic formula to solve that, then use a calculator to solve for**one**value of . I don't know if you have learned the graph of cosine or use the "unit circle" but you should know that there are**two**values of , between 0 and 360, such that . (That is, if y is between -1 and 1. If not, there is no solution.) There are either 0, or 2, or 4 solutions to this equation between 0 and 360. - April 1st 2013, 06:00 AMKrisltonRe: 2 cos \theta - 4 sin^2 \theta = 0
Yeah, I was trying to get it to a quadratic but I don't know my trig identities well enough to convert the sin to cos. I really appreciate your help, and I spend so many hours a day on math trying to understand, just sometimes I really struggle to see a link between identities. I'm sure I'll understand one day lol.