MHB 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:

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

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

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:

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

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

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?

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

 

[MarkFL]

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

No, let's start with:

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

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

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

Arrange in standard form:

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

Now factor...what do you get?

 

[jevanuD]

No, let's start with:

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

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

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

Arrange in standard form:

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

Now factor...what do you get?

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

 

[MarkFL]

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

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

$$3x^2-2x-1=0$$

 

[jevanuD]

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

$$3x^2-2x-1=0$$

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

 

[MarkFL]

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

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

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

From this, of course, we get:

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

$$\sin(\theta)=1$$

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:

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

From this, of course, we get:

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

$$\sin(\theta)=1$$

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: $${0}^{0}\le\theta\le360^{0}$$

 

[MarkFL]

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

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

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 $$y=-\frac{1}{3}$$...
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:

$$\theta=90^{\circ}$$

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:

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

Does that make sense?

 

[jevanuD]

Yes, and we should know immediately that:

$$\theta=90^{\circ}$$

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:

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

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:

$$\theta=360^{\circ}-\beta$$

$$\theta=180^{\circ}+\beta$$

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:

$$\theta=360^{\circ}-\beta$$

$$\theta=180^{\circ}+\beta$$

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:

$$\theta=360^{\circ}-\beta$$

Do you also see that:

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

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