Double / Half & Area Trig Help

**dondonlouie** · April 6th 2011, 12:51 AM

I have two questions:
1. Given that sin 3pi/10 = sqrt(5)+1/4 find an exact expression for cos (3pi/5)
I used cos(2theta) = 1 - 2sin^2(theta)
cos(3pi/5) = 1- 2sin^2(3pi/10)
= 1 - 2(sqrt(5)+1/4)^2
= 1- 2(3+sqrt(5)/8) = 4-3+sqrt(5)/4 = 1+sqrt(5)/4 <- i got that as the final answer but the answer key says that it's 1-sqrt(5)/4

2. What is the area of the triangle whose sides all have length r
i used A = 1/2absin(theta)
= 1/2 r^2sin(theta)
i got to that part and now i'm confused as to what to do next. the answer is sqrt(3)r^2/4
would it be right if i said that sin(theta) = 1/2 = pi/6 and then cos(pi/6) = sqrt(3)/2 and then use that to get sqrt(3)r^2/4?

**Prove It** · April 6th 2011, 02:23 AM

Do you mean $\displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}$ ?

You should know that $\displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}$ . Here $\displaystyle \theta = \frac{3\pi}{10}$ .

**dondonlouie** · April 6th 2011, 02:32 AM

Originally Posted by Prove It

Do you mean $\displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}$ ?

You should know that $\displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}$ . Here $\displaystyle \theta = \frac{3\pi}{10}$ .

yes i know but i still can't figure out what i did wrong to get 1+sqrt(5) instead of 1 - sqrt(5)

**Prove It** · April 6th 2011, 02:40 AM

Let's see...

$\displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2$

$\displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)$

$\displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}$

$\displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}$

$\displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}$

$\displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}$ . Go from here.

It appears you did not subtract ALL of the second term...

**dondonlouie** · April 6th 2011, 02:50 AM

Originally Posted by Prove It

Let's see...

$\displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2$

$\displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)$

$\displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}$

$\displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}$

$\displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}$

$\displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}$ . Go from here.

It appears you did not subtract ALL of the second term...

yay i got it~ wow, i didn't multiply the 8 to get 8-(6+2sqrt(5))

**Soroban** · April 6th 2011, 04:34 AM

Hello, dondonlouie!

$\text{2. What is the area of the triangle whose sides all have length } r\,\text{?}$

$\text{i used: }\:A \:=\: \frac{1}{2}ab\sin\theta \:=\:\frac{1}{2}r^2\sin\theta$

$\text{i got to that part and now i'm confused as to what to do next.}$

Since the triangle is equilateral, . $\theta \,=\,\frac{\pi}{3}$

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