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!

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- May 2nd 2012, 01:06 PMjonathanraxaPre-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 PMProve ItRe: Pre-Calc Homework Question
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 PMjonathanraxaRe: 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 PMProve ItRe: Pre-Calc Homework Question
- May 3rd 2012, 07:27 PMjonathanraxaRe: Pre-Calc Homework Question
I'm plugging

**sin(18degrees)**in for**t**in the equation. - May 3rd 2012, 08:44 PMjonathanraxaRe: Pre-Calc Homework Question
Attachment 23777

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