Trig

[Spectre927]

I'm trying to figure all solutions to the equation 2sin@=cos@ in the interval 0-360.
So do I need to switch cos@ to its pythag. identity or is there something simple I just keep overlooking. Thanks

 

[Spectre927]

2sin@= +/- S.R. 1-sin(squared)@ and I got it down to sin@= S.R. (1/3)...is that right?

 

[Spectre927]

S.R.1/5 2sin(26.565) = cos(26.565) and 2sin(206.565) = cos(206.565)

Look right? They equal on my calc. Its been 5 years since I took precalc and Im taking it again so IM trying to remember it...no more trig after my test in the morning though.

 

[soroban]

Hello, Spectre927!

I'm trying to figure all solutions to the equation \(\displaystyle 2\cdot\sin\theta\,=\,\cos\theta\) in the interval \(\displaystyle [0^o,\,360^o}\).

There is a simple method . . .

We have: \(\displaystyle \:2\cdot\sin\theta\,=\,\cos\theta\)

Divide by \(\displaystyle 2\cdot\cos\theta:\;\;\frac{2\cdot\sin\theta}{2\cdot\cos\theta}\:=\:\frac{\cos\theta}{2\cdot\cos\theta}\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{1}{2}\)

Hence: \(\displaystyle \,\theta\:=\:\tan^{-1}\left(\frac{1}{2}\right) \:\approx\:26.6^o,\:206.6^o\)

 

[Spectre927]

Hello, and thank you for that. I appreciate it