Is \(\theta=360^\circ-\beta\) When \(\sin(\beta)=\frac{1}{3}\)?

[jevanuD]

View attachment 8018

 

[S.G. Janssens]

Do you have ideas yourself? What standard trigonometric identities that you already know could possibly be applied here?

 

[jevanuD]

I am aware that $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and the other fundamental which is $1-\cos ^2(x)=\sin ^2(x)$ i just need some clarity on how its proven.x

 

[MarkFL]

I am aware that [M]\tan\left({}\right)[/M] = \sin\left({}\right)/\cos\left({}\right) and the other fundamental which is [M]\:1-\cos ^2\left(x\right)=\sin ^2\left(x\right)[/M] i just need some clarity on how its proven.

Use the $\sum$ button to wrap your code, instead of the M button. ;)

The two identities you cited:

\(\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\)

\(\displaystyle 1-\cos^2(\theta)=\sin^2(\theta)\)

will be very useful for the first problem...can you use them to make some substitutions?

 

[Greg]

Hi jevanuD. I've edited your second post to allow the $\LaTeX$ code to render properly and made some other minor edits. The delimiters for inline $\LaTeX$ are \$...\$. You can view the code by right-clicking on the $\LaTeX$ and selecting Show Math As > TeX Commands.

 

[jevanuD]

Use the $\sum$ button to wrap your code, instead of the M button. ;)

The two identities you cited:

\(\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\)

\(\displaystyle 1-\cos^2(\theta)=\sin^2(\theta)\)

will be very useful for the first problem...can you use them to make some substitutions?

Yes correct, I've made my substitutions to prove that question A is indeed true, however i am not sure how to start "B"

 

[MarkFL]

Yes correct, I've made my substitutions to prove that question A is indeed true, however i am not sure how to start "B"

For the first part of part (b), multiply through by $\sin(\theta)$ to get a quadratic in $\sin(\theta)$. Then factor...what do you get?

 

[jevanuD]

For the first part of part (b), multiply through by $\sin(\theta)$ to get a quadratic in $\sin(\theta)$. Then factor...what do you get?

\(\displaystyle 3\sin\left({\theta}\right){}^{2}=3\)?

 

[MarkFL]

\(\displaystyle 3\sin\left({\theta}\right){}^{2}=3\)?

No, let's start with:

\(\displaystyle 3\sin(\theta)=2+\frac{1}{\sin(\theta)}\)

Multiply through by $\sin(\theta)$ to get:

\(\displaystyle 3\sin^2(\theta)=2\sin(\theta)+1\)

Arrange in standard form:

\(\displaystyle 3\sin^2(\theta)-2\sin(\theta)-1=0\)

Now factor...what do you get?

 

[jevanuD]

No, let's start with:

\(\displaystyle 3\sin(\theta)=2+\frac{1}{\sin(\theta)}\)

Multiply through by $\sin(\theta)$ to get:

\(\displaystyle 3\sin^2(\theta)=2\sin(\theta)+1\)

Arrange in standard form:

\(\displaystyle 3\sin^2(\theta)-2\sin(\theta)-1=0\)

Now factor...what do you get?

\(\displaystyle \sin\left({\theta}\right)\left(3\sin\left({\theta}\right)-3\right)=0\)

 

[MarkFL]

\(\displaystyle \sin\left({\theta}\right)\left(3\sin\left({\theta}\right)-3\right)=0\)

No...what if I ask you to factor:

\(\displaystyle 3x^2-2x-1=0\)

 

[jevanuD]

No...what if I ask you to factor:

\(\displaystyle 3x^2-2x-1=0\)

\(\displaystyle \left(3x+1\right)\left(x-1\right)\)

 

[MarkFL]

\(\displaystyle \left(3x+1\right)\left(x-1\right)\)

Excellent...the same applies whether you have the quadratic in $x$ or $\sin(\theta)$ or any other expression. So, this gives you:

\(\displaystyle \left(3\sin(\theta)+1\right)\left(\sin(\theta)-1\right)=0\)

From this, of course, we get:

\(\displaystyle \sin(\theta)=-\frac{1}{3}\)

\(\displaystyle \sin(\theta)=1\)

Can you see how many solutions you are going to have, and in which quadrants?

 

[jevanuD]

Excellent...the same applies whether you have the quadratic in $x$ or $\sin(\theta)$ or any other expression. So, this gives you:

\(\displaystyle \left(3\sin(\theta)+1\right)\left(\sin(\theta)-1\right)=0\)

From this, of course, we get:

\(\displaystyle \sin(\theta)=-\frac{1}{3}\)

\(\displaystyle \sin(\theta)=1\)

Can you see how many solutions you are going to have, and in which quadrants?

Are you saying I should plot my 2 solutions into: \(\displaystyle {0}^{0}\le\theta\le360^{0}\)

 

[MarkFL]

Are you saying I should plot my 2 solutions into: \(\displaystyle {0}^{0}\le\theta\le360^{0}\)

What I would do is plot the unit circle, and the lines $y=1$ and \(\displaystyle y=-\frac{1}{3}\)...

View attachment 8024

From this graph, can you see how many solutions we're going to have on the required interval?

 

[jevanuD]

What I would do is plot the unit circle, and the lines $y=1$ and \(\displaystyle y=-\frac{1}{3}\)...
From this graph, can you see how many solutions we're going to have on the required interval?

from that I'm seeing 3 points marked, which is 3 solutions?

 

[MarkFL]

from that I'm seeing 3 points marked, which is 3 solutions?

Yes, and we should know immediately that:

\(\displaystyle \theta=90^{\circ}\)

is a solution corresponding to the intersection with the circle and the line $y=1$. Now, for the other two solutions, in quadrants III and IV, there are several ways you could approach finding them, and what I would do is observe they are symmetric about $270^{\circ}$...and so I would write:

\(\displaystyle \theta=\left(\frac{3\pi}{2}\pm\arccos\left(\frac{1}{3}\right)\right)\frac{180^{\circ}}{\pi}\)

Does that make sense?

 

[jevanuD]

Yes, and we should know immediately that:

\(\displaystyle \theta=90^{\circ}\)

is a solution corresponding to the intersection with the circle and the line $y=1$. Now, for the other two solutions, in quadrants III and IV, there are several ways you could approach finding them, and what I would do is observe they are symmetric about $270^{\circ}$...and so I would write:

\(\displaystyle \theta=\left(\frac{3\pi}{2}\pm\arccos\left(\frac{1}{3}\right)\right)\frac{180^{\circ}}{\pi}\)

Does that make sense?

wow I am lost there, but i would then solve this equation?

 

[MarkFL]

wow I am lost there, but i would then solve this equation?

Do you see that the quadrant III and IV solutions are symmetric about $270^{\circ}$?

 

[MarkFL]

Another approach we could use is to observe that the two solutions will be of the form:

\(\displaystyle \theta=360^{\circ}-\beta\)

\(\displaystyle \theta=180^{\circ}+\beta\)

Do you see what the reference angle $\beta$ must be?

 

[jevanuD]

Another approach we could use is to observe that the two solutions will be of the form:

\(\displaystyle \theta=360^{\circ}-\beta\)

\(\displaystyle \theta=180^{\circ}+\beta\)

Do you see what the reference angle $\beta$ must be?

We've only stated 2 angles 90 and 270, is it either of those?

 

[MarkFL]

We've only stated 2 angles 90 and 270, is it either of those?

Let's look at the following diagram:

View attachment 8029

Do you see that one of the solutions, the quadrant IV solution must be:

\(\displaystyle \theta=360^{\circ}-\beta\)

Do you also see that:

\(\displaystyle \sin(\beta)=\frac{1}{3}\)

If you see the above, then what is $\beta$?