+0

0
1
72
3
+22

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
+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
+22
0

Wait so I simplified it and I got (rounded to the nearest hundred) $$sin(\theta+ \beta)=-0.99$$

Sarcasticcarma  Oct 28, 2022
#3
+118141
+1

I see you have asked the same question again.

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