Trig equation (2)

Daniel123 · 2007-10-21 06:08:34

Thanks for the last one mathsy.. I have another my head is too sore to do:

tanθ = cosθ

thanks.

mathsyperson · 2007-10-21 06:14:58

That looks like it's going to be easy, but it's actually far less trivial than it seems. Doable though.

tanθ = cosθ
sinθ/cosθ = cosθ
sinθ = cos²θ
sinθ = 1 - sin²θ
sin²θ + sinθ - 1 = 0

And from there you can substitute to make a quadratic and then substitute back, as before.

Daniel123 · 2007-10-21 06:27:07

aah thanks.. i get this now!

Just out of interest... do you have to do the substitution?
I did:

sin²θ + sinθ - 1 = 0
(sinθ + 0.5)² - 1.25 = 0
sinθ = -0.5 ± √1.25
∴ θ = 38.2°

mathsyperson · 2007-10-21 06:59:41

Ah, fair point. I just got it in quadratic form and then stopped thinking.

Daniel123 · 2007-10-21 07:03:20

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