# Math Help - Sin(y)=Cos(y/3)

1. ## Sin(y)=Cos(y/3)

Hello everyone, I came across an old trigonometry problem which I'm not able to solve again and it's driving me crazy, here it is:

find the value of tan(x+y) knowing that:
tan(x)=-sqrt(2)/2
sin(y)=cos(y/3) with 19pi/4<=y<=5pi ( y in [19pi/4;5pi] ).

The final answer should be tan (x+y)= sqrt(2)-3.

My only problem is that I've a really hard time to find y. I tried to add pi/2 on both side to have cos or sin on both sides but it didn't help then I tried with a variable, pi/3=a => sin(3a)=cos(a) => sin(a)(4cosˆ2(a)-1)=cos(a) but then again it led me to nothing...
Can you please explain to me how to find y.

( I did the problem backwards and you should fin tan(y)=1-sqrt(2)).

2. Hello, sunmalus!

Find the value of $\tan(x+y)$

knowing that:
. . $\tan x \:=\:-\frac{\sqrt{2}}{2}$

. . $\sin y\:=\:\cos\tfrac{y}{2}\;\;\text{ where }\tfrac{19\pi}{4}\:\leq\: y\:\leq\: 5\pi$ . . ${\color{blue}y}$

The final answer should be: $\tan (x+y)\:=\: \sqrt{2}-3$ . .
I'm not getting this

We have: . $\sin y \;=\;\cos\tfrac{y}{2}$

Then:. . . . $\sin y \;=\;\pm\sqrt{\frac{1 + \cos y}{2}}$

Square: . $\sin^2\!y \;=\;\frac{1 + \cos y}{2}$

Then: . $1-\cos^2\!y \;=\;\frac{1+\cos y}{2} \quad\Rightarrow\quad 2\cos^2\!y + \cos y - 1 \;=\;0$

Factor: . $(2\cos y - 1)(\cos y + 1) \:=\:0$

Then: . $\begin{array}{ccccccccc}\cos y \;=\; \frac{1}{2} & \Rightarrow & y \;=\; \pm\frac{\pi}{3} & \text{not in Quadrant 2} \\ \\[-3mm] \cos y \;=\; \text{-}1 & \Rightarrow & \boxed{y \;=\; \pi} \end{array}$

And we have: . $\tan(x + \pi) \;=\;\tan x \;=\;-\frac{\sqrt{2}}{2}$

What's going on?

3. Originally Posted by sunmalus
Hello everyone, I came across an old trigonometry problem which I'm not able to solve again and it's driving me crazy, here it is:

find the value of tan(x+y) knowing that:
tan(x)=-sqrt(2)/2
sin(y)=cos(y/2) with 19pi/4<=y<=5pi ( y in [19pi/4;5pi] ).

The final answer should be tan (x+y)= sqrt(2)-3.

My only problem is that I've a really hard time to find y. I tried to add pi/2 on both side to have cos or sin on both sides but it didn't help then I tried with a variable, pi/3=a => sin(3a)=cos(a) => sin(a)(4cosˆ2(a)-1)=cos(a) but then again it led me to nothing...
Can you please explain to me how to find y.

( I did the problem backwards and you should fin tan(y)=1-sqrt(2)).

Does $\sin{y} = \cos{\frac{y}{2}}$ or does $\sin{y} = \cos{\frac{y}{3}}$?

4. ## correction

I am very very sorry, it's cos(y/3), I was a bit tired when I posted this. But thanks to Soroban I think I got it now.

Again a big sorry and many thanks!!

5. I spoke too fast, I'm still stuck...

6. Whew! that was a little lengthy!
Here's how I found y >>>>

$
sin(y) = cos(y/3)$
[1]

$sin(y) = 3sin(y/3) - 4sin^3(y/3)
$

and..
$cos(y/3) = sqrt(1-sin^2(y/3))$
so sub into [1]..
$(3sin(y/3) - 4sin^3(y/3))^2 = 1-sin(y/3)$
$9sin^2(y/3) + 16sin^6(y/3) - 24sin^4(y/3) = 1-sin^2(y/3)
$

same as..

$(1-sin^2(y/3))(8sin^2(y/3))(1-2sin^2(y/3)) + sin^2(y/3) = 1 -sin^2(y/3)
$

so..
$8sin^4(y/3)-8sin^2(y/3)+1 = 0$
$
sin^2(y/3) = 1/2 + sqrt(32)/16$

solution we want is -9pi/8
move y in positive direction to between 19pi/4 and 5pi there's your y

-9pi/8 + 3(2pi) = 39pi/8

7. ## Thanks

Thanks a lot to all of you!

8. RJH, I've a last question for you. How did you proceed to find that:

$
9sin^2(y/3) + 16sin^6(y/3) - 24sin^4(y/3) = 1-sin^2(y/3)
$

=>

$
(1-sin^2(y/3))(8sin^2(y/3))(1-2sin^2(y/3)) + sin^2(y/3) = 1 -sin^2(y/3)
$

because I looked how you started and I wanted to try to do the rest by myself but I got stuck at that step...

Do you have a methode or something, I know that you want to find something with the same power and I was looking for this, but by just "randomly" trying to find what it could be it would have taken me a while...

So any advice if I come across something like this again?

9. Hi

I propose you an alternative way

$\sin y = \cos \left(\frac{y}{3}\right)$ is equivalent to

$\cos \left(\frac{\pi}{2}-y\right) = \cos \left(\frac{y}{3}\right)$

$\cos a = \cos b$ leads to $a = b + 2k \pi$ or $a = -b + 2k \pi$

Therefore
$\frac{\pi}{2}-y = \frac{y}{3} + 2k \pi$ or $\frac{\pi}{2}-y = -\frac{y}{3} + 2k \pi$

$y = \frac{3\pi}{8} + 3k \frac{\pi}{2}$ or $y = \frac{3\pi}{4} + 3k \pi$

10. Thank you to running-gag!
I knew there must be a simpler way to do it...
sunmalus, I highly recommend that you follow running-gag's reasoning, my way is far too long winded