principle angles and related acute angles

**ahmedb** · December 15th 2011, 07:29 PM

how do you find them?
examples:
tan18
sin205
cos-55

**Prove It** · December 15th 2011, 09:08 PM

Originally Posted by ahmedb

how do you find them?
examples:
tan18
sin205
cos-55

I'm assuming these are in degrees.

$\displaystyle \begin{align*} \sin{(5x)} &= 16\sin^5{(x)} - 20\sin^3{(x)} + 5\sin{(x)} \\ \sin{(5\cdot 18^{\circ})} &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} \\ 1 &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} \\ 0 &= 16\sin^5{(18^{\circ})} - 20\sin^3{(18^{\circ})} + 5\sin{(18^{\circ})} - 1 \\ 0 &= 16y^5 - 20y^3 + 5y - 1 \textrm{ if we let }y = \sin{(18^{\circ})} \\ 0 &= (y - 1)\left(4y^2 + 2y - 1\right)^2 \\ 0 &= 4y^2 + 2y - 1 \\ 0 &= y^2 + \frac{1}{2}y - \frac{1}{4} \\ 0 &= y^2 + \frac{1}{2}y + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - \frac{1}{4} \\ 0 &= \left( y + \frac{1}{4}\right)^2 - \frac{5}{16} \\ \frac{5}{16} &= \left(y + \frac{1}{4}\right)^2 \\ \pm \frac{\sqrt{5}}{4} &= y + \frac{1}{4} \\ \frac{-1 \pm \sqrt{5}}{4} &= y \end{align*}$

With a little checking, you'll see that $\displaystyle \begin{align*} \sin{(18^{\circ})} = \frac{-1 + \sqrt{5}}{4} \end{align*}$ is the correct solution.

Now use a similar method using

$\displaystyle \begin{align*} \cos{(5x)} = 16\cos^5{(x)} - 20\cos^3{(x)} + 5\cos{(x)} \end{align*}$ to get the value of $\displaystyle \begin{align*} \cos{(18^{\circ})} \end{align*}$ .

Then you can find $\displaystyle \begin{align*} \tan{(18^{\circ})} \end{align*}$ using $\displaystyle \begin{align*} \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}} \end{align*}$ .

**ahmedb** · December 18th 2011, 07:08 PM

what i mean is like quadrant 1, 2, 3, and 4.

**sbhatnagar** · December 19th 2011, 05:13 AM

Here's a little easier method:

$\begin{align*} x &= 18^{\circ} \\ 5x &= 90^{\circ} \\ 2x+ 3x &= 90^{\circ} \\ 2x &= 90^{\circ} -3x \\ \sin(2x)&= \sin(90^{\circ}-3x) \\ \sin(2x) &= \cos(3x) \\ 2\sin(x)\cos(x) &= \cos^3(x)-3\sin^2(x)\cos(x) \\ 2\sin(x) &= \cos^2(x)-3\sin^2(x) \\ 2\sin(x) &= 1-\sin^2(x)-3\sin^2(x) \\ 4\sin^2(x)+2\sin(x)-1 &= 0 \\ \sin(x) &= \frac{-1 \pm \sqrt{5}}{4} \quad [\text{Quadratic Formula}] \\ \sin(18^{\circ}) &= \frac{-1 + \sqrt{5}}{4} \end{align*}$

$\sin(205 ^{\circ})=-\sin(25 ^{\circ} )$

$\cos(-55 ^{\circ}) = \cos(55 ^{\circ})$

Unfortunately, $\sin(25 ^{\circ} )$ and $\cos(55 ^{\circ})$ have no exact values.

**skeeter** · December 19th 2011, 06:40 AM

Originally Posted by ahmedb

what i mean is like quadrant 1, 2, 3, and 4.

... are you asking about reference angles?

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