# Complex Question (1 Viewer)

#### hmim

##### Member
for the question 32x^5 - 40x^3 +10x -1 = 0 (leave your answer in terms of sin theta:

In this solution, i'm confused why 49pi/30 and 53pi/30 aren't included as part of the values of x?

#### tywebb

##### dangerman
There have to be only 5 distinct answers. sin(49pi/30) and sin(53pi/30) are included because

$\bg_white \sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}$ and

$\bg_white \sin\frac{53}{30}=\sin\frac{37\pi}{30}$

So it matters not if you list them as

$\bg_white \sin\frac{\pi}{30},\sin\frac{\pi}{6},\sin\frac{13\pi}{30},\sin\frac{37\pi}{30},\sin\frac{41\pi}{30}$ or

$\bg_white \sin\frac{\pi}{30},\sin\frac{\pi}{6},\sin\frac{13\pi}{30},\sin\frac{53\pi}{30},\sin\frac{49\pi}{30}$

but to keep them distinct for example you can't have $\bg_white \sin\frac{49\pi}{30}$ and $\bg_white \sin\frac{41\pi}{30}$ in the list because they are the same.

I just think they chose the smallest values of $\bg_white \theta$ to use in the list.

A simpler example would be to list $\bg_white \sin\frac{\pi}{6}$ instead of $\bg_white \sin\frac{5\pi}{6}$

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#### hmim

##### Member
There have to be only 5 distinct answers. sin(49pi/30) and sin(53pi/30) are included because

$\bg_white \sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}$ and

$\bg_white \sin\frac{53}{30}=\sin\frac{37\pi}{30}$

So it matters not if you list them as

$\bg_white \sin\frac{\pi}{30},\sin\frac{\pi}{6},\sin\frac{13\pi}{30},\sin\frac{37\pi}{30},\sin\frac{41\pi}{30}$ or

$\bg_white \sin\frac{\pi}{30},\sin\frac{\pi}{6},\sin\frac{13\pi}{30},\sin\frac{53\pi}{30},\sin\frac{49\pi}{30}$

but to keep them distinct for example you can't have $\bg_white \sin\frac{49\pi}{30}$ and $\bg_white \sin\frac{41\pi}{30}$ in the list because they are the same.

I just think they chose the smallest values of $\bg_white \theta$ to use in the list.

A simpler example would be to list $\bg_white \sin\frac{\pi}{6}$ instead of $\bg_white \sin\frac{5\pi}{6}$
Hmm, how did you find that
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}&hash=0720ef0599462a62c3e93411a22bf94a$
and

$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{53}{30}=\sin\frac{37\pi}{30}&hash=a9a542a5976afa31ae7154f1cc3828c4$
??

I'm a bit confused. I know its equal but how did you get it?
Any help would be appreciated, thanks

#### tywebb

##### dangerman
3 ways.

1. Check with calculator

$\bg_white \sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=0$ and $\bg_white \sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=0$

2. Use sum to product

$\bg_white \!\!\!\!\!\!\sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=\sin\left(\frac{3\pi}{2}+\frac{2\pi}{15}\right)-\sin\left(\frac{3\pi}{2}-\frac{2\pi}{15}\right)=2\cos\frac{3\pi}{2}\sin\frac{2\pi}{15}=0$ and

$\bg_white \!\!\!\!\!\!\sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=\sin\left(\frac{3\pi}{2}+\frac{4\pi}{15}\right)-\sin\left(\frac{3\pi}{2}-\frac{4\pi}{15}\right)=2\cos\frac{3\pi}{2}\sin\frac{4\pi}{15}=0$

3. Use symmetry

$\bg_white \sin\frac{49\pi}{30}=\sin\left(\pi-\frac{49\pi}{30}\right)=\sin\left(-\frac{19\pi}{30}\right)=\sin\left(2\pi-\frac{19\pi}{30}\right)=\sin\frac{41\pi}{30}$ and

$\bg_white \sin\frac{53\pi}{30}=\sin\left(\pi-\frac{53\pi}{30}\right)=\sin\left(-\frac{23\pi}{30}\right)=\sin\left(2\pi-\frac{23\pi}{30}\right)=\sin\frac{37\pi}{30}$

#### synthesisFR

##### Well-Known Member
Hmm, how did you find that
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}&hash=0720ef0599462a62c3e93411a22bf94a$
and

$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{53}{30}=\sin\frac{37\pi}{30}&hash=a9a542a5976afa31ae7154f1cc3828c4$
??

I'm a bit confused. I know its equal but how did you get it?
Any help would be appreciated, thanks
for these questions i was taught to always draw the unit circle
so for example this:
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}&hash=0720ef0599462a62c3e93411a22bf94a$
and
sin 45pi/30 would be the negative y axis bc its 3pi/2, with 49pi/30 and 41pi/30 equally spaced on each side. Sin is negative in 3rd and quadrant, and both these angles represent the same length bc its equally spaced from sin45pi/30, so they would give the same values, so u cancel one out.
Also for exams i was told to always use the smaller angle if it repeats so u would take 41pi/30. lmk if u need a diagram

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#### hmim

##### Member
for these questions i was taught to always draw the unit circle
so for example this:
$image=https://latex.codecogs.com/png.latex?\bg_white%20\sin\frac{49\pi}{30}=\sin\frac{41\pi}{30}&hash=0720ef0599462a62c3e93411a22bf94a$
and
sin 45pi/30 would be the negative y axis bc its 3pi/2, with 49pi/30 and 41pi/30 equally spaced on each side. Sin is negative in 3rd and quadrant, and both these angles represent the same length bc its equally spaced from sin45pi/30, so they would give the same values, so u cancel one out.
Also for exams i was told to always use the smaller angle if it repeats so u would take 41pi/30. lmk if u need a diagram
Ohh thanks, that makes sense. It would be great if you could provide the diagram though!

#### hmim

##### Member
3 ways.

1. Check with calculator

$\bg_white \sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=0$ and $\bg_white \sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=0$

2. Use sum to product

$\bg_white \!\!\!\!\!\!\sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=\sin\left(\frac{3\pi}{2}+\frac{2\pi}{15}\right)-\sin\left(\frac{3\pi}{2}-\frac{2\pi}{15}\right)=2\cos\frac{3\pi}{2}\sin\frac{2\pi}{15}=0$ and

$\bg_white \!\!\!\!\!\!\sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=\sin\left(\frac{3\pi}{2}+\frac{4\pi}{15}\right)-\sin\left(\frac{3\pi}{2}-\frac{4\pi}{15}\right)=2\cos\frac{3\pi}{2}\sin\frac{4\pi}{15}=0$

3. Use symmetry

$\bg_white \sin\frac{49\pi}{30}=\sin\left(\pi-\frac{49\pi}{30}\right)=\sin\left(-\frac{19\pi}{30}\right)=\sin\left(2\pi-\frac{19\pi}{30}\right)=\sin\frac{41\pi}{30}$ and

$\bg_white \sin\frac{53\pi}{30}=\sin\left(\pi-\frac{53\pi}{30}\right)=\sin\left(-\frac{23\pi}{30}\right)=\sin\left(2\pi-\frac{23\pi}{30}\right)=\sin\frac{37\pi}{30}$
Thanks! Method 3 looks useful

#### Drongoski

##### Well-Known Member
Another way to look at it:

$\bg_white \frac{49\pi}{30} = 2\pi - \frac{11\pi}{30} and \frac {41\pi}{30} = \pi + \frac {11\pi}{30}\\ \\ So: sin \frac {49\pi}{30} and sin \frac {41\pi}{30}$
have the same related angles (11pi/30) and they are both negative, being in the 3rd & 4th quadrants resp. Similarly:

$\bg_white \frac {53\pi}{30} = 2\pi - \frac {7\pi}{30} and \frac {37\pi}{30} = \pi+ \frac {7\pi}{30} \\ \\
both have equal related angles 7pi/30 and both have sine in the negative quadrants .$

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