# Pre-Calc Homework Question

• May 2nd 2012, 01:06 PM
jonathanraxa
Pre-Calc Homework Question
A problem from my homework:

Explain why sin(18degrees)=cos(72degrees). Then using the previous problem, explain why:

(8t^4)-(8t^2)-t+1=0

Thank You!
• May 2nd 2012, 02:27 PM
Prove It
Re: Pre-Calc Homework Question
Quote:

Originally Posted by jonathanraxa
A problem from my homework:

Explain why sin(18degrees)=cos(72degrees). Then using the previous problem, explain why:

(8t^4)-(8t^2)-t+1=0

Thank You!

Draw any right-angle triangle. Call one of the non-right angles \displaystyle \displaystyle \begin{align*} \theta^{\circ} \end{align*}, then the other angle is \displaystyle \displaystyle \begin{align*} 90^{\circ} - \theta^{\circ} \end{align*}. Do you understand why? The three angles in a triangle have to add to \displaystyle \displaystyle \begin{align*} 180^{\circ} \end{align*}.

Now note that from the point of view of each angle, the hypotenuses are the same, but the opposite and adjacent sides switch.

So \displaystyle \displaystyle \begin{align*} \sin{\left(90^{\circ} - \theta^{\circ}\right)} = \cos{\theta^{\circ}} \end{align*} and \displaystyle \displaystyle \begin{align*} \cos{\left(90^{\circ} - \theta^{\circ}\right)} = \sin{\theta^{\circ}} \end{align*}.
• May 3rd 2012, 01:07 PM
jonathanraxa
Re: Pre-Calc Homework Question
Ok I got that thanks. As for the second portion, given that:
t=sin(18degrees)
what are the steps in solving:
(8t^4)-(8t^2)-t+1=0 ?
• May 3rd 2012, 07:10 PM
Prove It
Re: Pre-Calc Homework Question
Quote:

Originally Posted by jonathanraxa
A problem from my homework:

Explain why sin(18degrees)=cos(72degrees). Then using the previous problem, explain why:

(8t^4)-(8t^2)-t+1=0

Thank You!

What is t being used to represent?
• May 3rd 2012, 07:27 PM
jonathanraxa
Re: Pre-Calc Homework Question
I'm plugging sin(18degrees) in for t in the equation.
• May 3rd 2012, 08:44 PM
jonathanraxa
Re: Pre-Calc Homework Question
Attachment 23777

I want to find out how this is true. I don't know where to begin =/