Results 1 to 6 of 6

Math Help - PreCal Identities and Formulas

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    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 2sinē(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 2cotē(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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    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}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,567
    Thanks
    1426
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,567
    Thanks
    1426
    Quote Originally Posted by SwaGGeReR View Post
    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 2sinē(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 2cotē(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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    2
    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!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,567
    Thanks
    1426
    Quote Originally Posted by SwaGGeReR View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 30th 2011, 09:50 PM
  2. precal 5.1 to 5.3
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 4th 2009, 04:25 AM
  3. Precal help
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: October 28th 2008, 04:08 PM
  4. precal help
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 15th 2008, 03:52 PM
  5. Replies: 1
    Last Post: November 14th 2007, 08:30 PM

Search Tags


/mathhelpforum @mathhelpforum