Using Angle Difference to Get Exact Value

[zolton5971]

Need some help with this problem.

Use a sum or difference formula to find the exact value of the following.

sin14π/15 cos11π/60 -cos14π/15 sin11π/60=

Thanks

 

[MarkFL]

Hello and welcome to MHB, zolton5971! :D

For future reference, we ask those posting questions to show what they have tried so our helpers know where you are stuck or where you may have gone astray.

Now, the formula we want to bring to bear here is:

[box=blue]

Angle Sum/Difference Formula for Sine​

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\(\displaystyle \sin(\alpha\pm\beta)=\sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta)\tag{1}\)[/box]

Can you identify $\alpha$ and $\beta$ in this case and apply the above formula (1)?

 

[zolton5971]

A is 14Pi/15 and B is 11pi/60

 

[MarkFL]

A is 14Pi/15 and B is 11pi/60

Okay, good! (Yes)

Now, what does the formula in (1) tell us about the original difference you stated?

 

[zolton5971]

I'm having a little trouble with that?

 

[MarkFL]

Well, taking the values for $\alpha$ and $\beta$ which you correctly identified, and plugging into (1), we obtain:

\(\displaystyle \sin\left(\frac{14\pi}{15}-\frac{11\pi}{60}\right)=\sin\left(\frac{14\pi}{15}\right)\cos\left(\frac{11\pi}{60}\right)-\cos\left(\frac{14\pi}{15}\right)\sin\left(\frac{11\pi}{60}\right)\)

Now, we have on the right side of this equation the expression originally given, and so we know we can find its value by using what's on the left. Can you simplify the angle which is the argument of the sine function there?

 

[zolton5971]

would the two cos(11π60)−cos(14π15) cancel each other out

 

[MarkFL]

would the two cos(11π60)−cos(14π15) cancel each other out

No, we have used the angle difference formula to simplify the original expression to:

\(\displaystyle \sin\left(\frac{14\pi}{15}-\frac{11\pi}{60}\right)\)

Now what you want to do is evaluate the difference representing the angle of the sine function. So, get a common denominator, and then reduce the result...what do you find?

 

[zolton5971]

For some reason I am getting .9847 is that right?

 

[MarkFL]

For some reason I am getting .9847 is that right?

No, and without seeing your work, I have no idea where you went astray. Why not follow my suggestion above, and tell me to what the argument for the sine function reduces?

edit: You will be able to get the exact value, not a decimal approximation.

 

[zolton5971]

Would you be able to show me how to do that?

 

[MarkFL]

We need to evaluate the expression:

\(\displaystyle \frac{14\pi}{15}-\frac{11\pi}{60}\)

What is the LCD?

 

[zolton5971]

is it 154?

 

[MarkFL]

is it 154?

No the LCD is the LCM of the two denominators, and since 60 is divisible by 15, it is 60, so this means that in order to get the term on the left to have a denominator of 60, we need to multiply it by 1 in the form of:

\(\displaystyle 1=\frac{60/15}{60/15}=\frac{4}{4}\)

So, we now have:

\(\displaystyle \frac{14\pi}{15}\cdot\frac{4}{4}-\frac{11\pi}{60}=\frac{56\pi}{60}-\frac{11\pi}{60}\)

Now we have a common denominator, so to what does the expression reduce?

 

[zolton5971]

3pi/4?

 

[MarkFL]

3pi/4?

Correct...so this means we know the original expression is equal to:

\(\displaystyle \sin\left(\frac{3\pi}{4}\right)\)

This is a special angle...what is the value of the above sine function?

 

[zolton5971]

is it 2.35?

 

[MarkFL]

is it 2.35?

No, recall that for any real number $x$, we must have:

\(\displaystyle -1\le\sin(x)\le1\)

You should know this value as one of your special angles, or by converting this quadrant II angle to a quadrant I angle using the identity:

\(\displaystyle \sin(\theta)=\sin(\pi-\theta)\)

So, we could also use:

\(\displaystyle \sin\left(\frac{\pi}{4}\right)\)

What is the value of this?

 

[zolton5971]

sqrt2/2?

 

[MarkFL]

sqrt2/2?

Yes, so we know the original expression has a value of:

\(\displaystyle \frac{\sqrt{2}}{2}\)

 

[zolton5971]

Thanks I really appreciate it!