PreCal Identities and Formulas

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April 26th 2009, 06:05 PM
SwaGGeReR

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.

Thanks in advance guys :)
April 26th 2009, 07:48 PM
Soroban

Hello, SwaGGeReR!

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

Here are a few of them . . .

Quote:

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.

Quote:

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

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

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

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

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

Quote:

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.

Quote:

$4.\;\;\tan(y)=\tfrac{3}{4}\:\text{ and }\;\sin(y) < 0$
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: . $\sin(2y) \;=\;2\sin(y)\cos(y) \;=\; 2\left(-\frac{3}{5}\right) \left(-\frac{4}{5}\right) \;=\;\frac{24}{25}$
April 26th 2009, 08:46 PM
Prove It

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.
April 26th 2009, 08:48 PM
Prove It

Quote:

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.

Thanks in advance guys :)

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?
April 26th 2009, 09:06 PM
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!
April 27th 2009, 12:50 AM
Prove It

Quote:

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.