# Trigonometry, solving for between 0 and 360

• Apr 14th 2008, 02:35 PM
Nick87
Trigonometry, solving for between 0 and 360
Hi!

I'm trying to solve θ between 0° and 360°

I know that if $\displaystyle Sin$θ = 0.7 then θ = 44.4 (because $\displaystyle sin^-1$(0.7)= 44.4 to 1.d.p)

I know this for $\displaystyle Cos$ and $\displaystyle Tan$ but how do I solve for Cosec, Cot and Sec?

In particular the questions im looking at are:

1) Cosecθ = 2.3
2) Cotθ = -4.26
3) Secθ = 1.5

Im not really looking for the answers but the method itself, thank you!
• Apr 14th 2008, 02:38 PM
Mathstud28
You could
Quote:

Originally Posted by Nick87
Hi!

I'm trying to solve θ between 0° and 360°

I know that if $\displaystyle Sin$θ = 0.7 then θ = 44.4 (because $\displaystyle sin^-1$(0.7)= 44.4 to 1.d.p)

I know this for $\displaystyle Cos$ and $\displaystyle Tan$ but how do I solve for Cosec, Cot and Sec?

In particular the questions im looking at are:

1) Cosecθ = 2.3
2) Cotθ = -4.26
3) Secθ = 1.5

Im not really looking for the answers but the method itself, thank you!

Do one of two things...just take the $\displaystyle arcsc,arcot,arsec$ or use the theorem that for he co-functions you just reciprocate the $\displaystyle \theta$...example $\displaystyle tan(\theta)=cot\bigg(\frac{1}{\theta}\bigg)$...also note the converse works $\displaystyle cot(\theta)=tan\bigg(\frac{1}{\theta}\bigg)$
• Apr 14th 2008, 02:39 PM
Mathstud28
Oh yeah
and $\displaystyle arcsec(x)=sec^{-1}(x)$ as for all of the arc things..if you were unfamiliar with that notation