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cosθ=−√2/6 , where π≤θ≤3π/2 .

tanβ=5/12 , where 0≤β≤π/2 .

What is the exact value of sin(θ+β)?

Enter your answer, as a single fraction in simplified form, in the box.

 

sin(θ+β)  = ?

 Oct 27, 2022
 #1
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+2

Hi Sarcasticcarma!

\(cos(\theta)=-\dfrac{\sqrt{2}}{6} \text{ where } \pi \le \theta \le \dfrac{3\pi}{2}\)

\(tan(\beta)=\dfrac{5}{12} \text{ where } 0 \le \beta \le \dfrac{\pi}{2}\)

\(\text{What is the exact value of } sin(\theta+\beta)?\)

For this question, we need to know the expansion of \(sin(A+B)\) in general.

So remember: \(sin(A+B)=sin(A)cos(B)+sin(B)cos(A)\)

So: \(sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta)\)                                                                                       (1)

To solve this question, we just need to find: \(sin(\theta),cos(\beta),sin(\beta),cos(\theta)\)  (But we already have \(cos(\theta)\)).

If \(cos(\theta)=-\dfrac{\sqrt{2}}{6}\), then we have two ways to get \(sin(\theta)\).

First way: Recall: \(cos^2(\theta)+sin^2(\theta)=1 \iff sin(\theta)=\pm \sqrt{1-cos^2(\theta)}\)

So: \(sin(\theta)=\pm \sqrt{1-\dfrac{{2}}{36}}=\pm\sqrt{\dfrac{17}{18}} \\ \) But shall we take the positive or negative?
We must look at the given interval of \(\theta\) which is between \(\pi \text{ and } \dfrac{3\pi}{2}\)

This is the third quadrant, so \(sin(\theta)\) is negative. (Determined by the "CAST" rule or drawing sine graph.)

Thus, \(sin(\theta)=-\sqrt{\dfrac{17}{18}}\)

The second way to get \(sin(\theta)\) is much faster and in fact, we will use it to get \(cos(\beta),sin(\beta)\), as follows:

Given: \(tan(\beta)=\dfrac{5}{12}\)

Then, draw a right angle triangle, and choose any angle to be \(\beta\)

Then, we know \(tan(\beta)=\dfrac{5}{12}\), so the opposite side to the angle you chose is 5 and the adjacent must be 12; and by pythagoras theorem, the hypotenuse is 13. (Draw it!)

Now what is \(sin(\beta)?,cos(\beta)?\) This is easily done via the triangle we already drew!
So: 

\(sin(\beta)=\dfrac{5}{13}\\ \\ \\ cos(\beta)=\dfrac{12}{13}\)

And since we are in the first quadrant, all of them will be positive.

So finally by (1):

\(sin(\theta+\beta)=sin(\theta)cos(\beta)+sin(\beta)cos(\theta) \\ sin(\theta+\beta)=-\sqrt{\dfrac{17}{18}}*\dfrac{12}{13}+\dfrac{5}{13}*(-\dfrac{\sqrt{2}}{6})\)

You can simplify it :).

I hope this helps!

 Oct 28, 2022
 #2
avatar+22 
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Wait so I simplified it and I got (rounded to the nearest hundred) \(sin(\theta+ \beta)=-0.99\)

Sarcasticcarma  Oct 28, 2022
 #3
avatar+118141 
+1

I see you have asked the same question again.

Please do not do that.

You can make another post just putting a link back to the original question and asking people to answer on the original if you want to.

 

Guest has put a lot of work into this answers (I have not looked at it myself but at a glance it looks impressive, thanks guest.

 

Don't you understand the answer given?  You were asked for an exact value, not an approximation. 

You didn't thank guest for his/her efforts either.  Didn't even give him a point.

 

I know you are a new member so some of these oversites could just be you not understanding how it works. I am sorry if I have sounded to harsh.

 

Please simplify guests answer and tell us what you get. I want to see if you can do that properly.

 

For you and others:

You have another answer from asinus here:

https://web2.0calc.com/questions/cos-2-6-where-3-2

I have not looked at either answer.

Melody  Oct 29, 2022

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