# Math Help - PreCal Identities and Formulas

1. ## PreCal Identities and Formulas

Our teacher says that this unit (identities and formulas) is the hardest, and I'm starting to believe him haha. Anyways, I need some assistance on a few problems (please excuse my poor English):

1. solve 2sin²(y)-5sin(y)=-3

I factored and found that sin(y)=1 and sin(y)=3/2. Because sin(y)=1, I concluded that one of the solutions must be pi/2. However, with the sin(y)=3/2, that is not a special triangle, so how do I find the solutions?

2. Verify: cos(y+pi)=-cos(y)

I'm completely stumped. I know that there is an identity cos((pi/2)-x)=sinx, but that's the closest I could find...

3. solve 2cot²(y)+3csc(y)=0

I factored and found csc(y)=-2 and csc(y)=-1. Therefore, I found the solution set to be (3pi)/2, (7pi)/6, and (11pi)/6. I plugged in the answers in the original problem to check, and while (7pi)/6 and (11pi)/6 work, (3pi)/2 doesn't. I derived the answer choice (3pi)/2 from the csc(y)=-1; if csc(y)=-1, then shouldn't sin(y) also equal -1, therefore making (3pi)/2 a valid answer?

4. tan(y)=3/4 and sin(y)<0, what is sin(2x)?

I know that this must be located in the quadrant III but that's as far as I've got. It's a 3,4,5 triangle, but I don't know the angle.

5. find the solutions in [0,2pi) for sin(2y)+sin(y)=0.

I was thinking that you might factor this to be sin(2y+1), but I'm not completely sure. I know that sin(2y) includes multiple angles but I can't seem to find the solution.

2. Hello, SwaGGeReR!

Your English is excellent ... and thank you for showing your work!

Here are a few of them . . .

1. Solve: . $2\sin^2\!y-5\sin y \:=\:-3$

I factored and found that: . $\sin y=1\,\text{ and }\,\sin y=\tfrac{3}{2}$ .
Good!
Because $\sin y=1$, I concluded that one solution is $\tfrac{\pi}{2}$ .
Yes!
However, with $\sin y=\tfrac{3}{2}$, that is not a special triangle, so how do I find the solutions?
Recall that: . $-1 \:\leq\: \sin\theta \:\leq \: 1$ . . . Hence: . $\sin y = \tfrac{3}{2}$ .has no solution.

2. Verify: . $\cos(y+\pi)\:=\:-\cos(y)$
Use the identity: . $\cos(A + B) \:=\:\cos(A)\cos(B) - \sin(A)\sin(B)$

We have: . $\cos(y + \pi) \;=\;\cos(y)\cos(\pi) - \sin(y)\sin(\pi)$

. . . . . . . . . . . . . . . $= \;\cos(y)\cdot(-1) - \sin(y)\cdot(0)$

. . . . . . . . . . . . . . . $= \quad -\cos(y)$

3. Solve: . $2\cot^2(y)+3\csc(y)\:=\:0$
You must have factored incorrectly . . .

. . $2[\csc^2(y) - 1] + 3\csc(y) \:=\:0 \quad\Rightarrow\quad 2\csc^2(y) + 3\csc(y) - 2 \:=\:0$

Factor: . $\bigg[\csc(y) + 2\bigg]\,\bigg[2\csc(y) - 1\bigg] \:=\:0$

We have: . $\csc(y) \:=\:-2 \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{7\pi}{6},\:\tfrac{11\pi}{6}}$

And: . $2\csc(y) - 1 \:=\:0 \quad\Rightarrow\quad \csc (y) \:=\:\frac{1}{2}$ . . . which has no solution.

$4.\;\;\tan(y)=\tfrac{3}{4}\:\text{ and }\;\sin(y) < 0$
What is $\sin(2y)$?

I know that $y$ is in quadrant III. .
Good!
It's a 3-4-5 triangle, but I don't know the angle. .
You don't need it.

In Quadrant III: . $\tan(y) \:=\:\frac{3}{4} \:=\:\frac{opp}{adj}$

So we have: . $opp = -3,\:adj = -4\quad\hdots\quad hyp = 5$

. . Hence: . $\sin(y) \:=\:-\frac{3}{5},\quad\cos(y) \:=\:-\frac{4}{5}$

Therefore: . $\sin(2y) \;=\;2\sin(y)\cos(y) \;=\; 2\left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \;=\;\frac{24}{25}$

3. Also note that for Question 1

$\sin{y} = 1 \implies y = \frac{\pi}{2} + 2\pi n, n \in \mathbf{Z}$

to account for all the rotations around the unit circle.

4. Originally Posted by SwaGGeReR
Our teacher says that this unit (identities and formulas) is the hardest, and I'm starting to believe him haha. Anyways, I need some assistance on a few problems (please excuse my poor English):

1. solve 2sin²(y)-5sin(y)=-3

I factored and found that sin(y)=1 and sin(y)=3/2. Because sin(y)=1, I concluded that one of the solutions must be pi/2. However, with the sin(y)=3/2, that is not a special triangle, so how do I find the solutions?

2. Verify: cos(y+pi)=-cos(y)

I'm completely stumped. I know that there is an identity cos((pi/2)-x)=sinx, but that's the closest I could find...

3. solve 2cot²(y)+3csc(y)=0

I factored and found csc(y)=-2 and csc(y)=-1. Therefore, I found the solution set to be (3pi)/2, (7pi)/6, and (11pi)/6. I plugged in the answers in the original problem to check, and while (7pi)/6 and (11pi)/6 work, (3pi)/2 doesn't. I derived the answer choice (3pi)/2 from the csc(y)=-1; if csc(y)=-1, then shouldn't sin(y) also equal -1, therefore making (3pi)/2 a valid answer?

4. tan(y)=3/4 and sin(y)<0, what is sin(2x)?

I know that this must be located in the quadrant III but that's as far as I've got. It's a 3,4,5 triangle, but I don't know the angle.

5. find the solutions in [0,2pi) for sin(2y)+sin(y)=0.

I was thinking that you might factor this to be sin(2y+1), but I'm not completely sure. I know that sin(2y) includes multiple angles but I can't seem to find the solution.

5. $\sin{(2y)} + \sin{y} = 0$.

Recall that $\sin{(2y)} = 2\sin{y}\cos{y}$

So $2\sin{y}\cos{y} + \sin{y} = 0$

$\sin{y}(2\cos{y} + 1) = 0$.

Can you go from here?

5. Thank you guys for the amazing help!

I've managed to solve all the problems but I'm curious about some things:

When the solution set is unrestricted, I know you must account for all the coterminal angles by adding 2npi. What does this signify though?

$n \in \mathbf{Z}$

And in problem number 5, I got sin(y)=0 and cos(y)=-1/2. The solutions I got were 0, pi, 2pi/3, 4pi/3. However, when plugging it into the calculator, only 0 and pi worked. Is that the only two viable answers for this problem?

Thanks!

6. Originally Posted by SwaGGeReR
Thank you guys for the amazing help!

I've managed to solve all the problems but I'm curious about some things:

When the solution set is unrestricted, I know you must account for all the coterminal angles by adding 2npi. What does this signify though?

$n \in \mathbf{Z}$

And in problem number 5, I got sin(y)=0 and cos(y)=-1/2. The solutions I got were 0, pi, 2pi/3, 4pi/3. However, when plugging it into the calculator, only 0 and pi worked. Is that the only two viable answers for this problem?

Thanks!
The set $\mathbf{Z}$ is the set of integers, in other words

$(\dots, -3, -2, -1, 0 , 1, 2, 3, \dots)$.

So if $n \in \mathbf{Z}$, this means that $n$ is an integer.

Therefore $2\pi n$, if $n$ is an integer, means every integer multiplied by $2\pi$.

For Q5. Did you enter the values into the calculator correctly? There is no reason why $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ shouldn't work.

Just check

$\sin{\left(2\times\frac{2\pi}{3}\right)} + \sin{\frac{2\pi}{3}} = \sin{\frac{4\pi}{3}} + \sin{\frac{2\pi}{3}}$

$= -\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$

$= 0$ as required.