Results 1 to 6 of 6

Math Help - Sum and difference identities

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    197

    Sum and difference identities part 2

    Question cos 165
    can it be solved by using the special angle
    cos 150 + cos 15
    cos 150 + cos (60-45)

    Can it be solved in this way and will it provide me the correct answer
    Last edited by mj.alawami; February 9th 2009 at 08:25 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by mj.alawami View Post
    Angles for sum and difference identities?

    how to find the angle of the sum and difference identities
    for example sin 15 =(60-45) form where did the 60 and 45 come from and how did you get them in order to solve the problem
    Why can't we use instead of 60 -45 >>>>>>10 +5?
    Us the formula:

     \sin(A-B) = \sin A \cos B - \cos A \sin B

    and see my attachment to see why we use 60 and 45 instead of 10 and 5. We can easily derive triangles that give us SOH CAH TOA expression for sin(30), sin(60), sin(45) and cos(30), cos(60) and cos(45).

    We can not do the same for the angles 10 and 5.
    Attached Thumbnails Attached Thumbnails Sum and difference identities-asd.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    197
    Quote Originally Posted by Mush View Post
    Us the formula:

     \sin(A-B) = \sin A \cos B - \cos A \sin B

    and see my attachment to see why we use 60 and 45 instead of 10 and 5. We can easily derive triangles that give us SOH CAH TOA expression for sin(30), sin(60), sin(45) and cos(30), cos(60) and cos(45).

    We can not do the same for the angles 10 and 5.
    Can you solve my newly edited question please
    Thank you very much for taking the time to help me
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by mj.alawami View Post
    Question cos 165
    can it be solved by using the special angle
    cos 150 + cos 15
    cos 150 + cos (60-45)

    Can it be solved in this way and will it provide me the correct answer
    Hrm not quite.

     \cos(165) = \cos(150+15) = \cos(150)\cos(15)-\sin(150)\sin(15)

     = \cos(90 + 60)\cos(60-45) - \sin(90+60) \sin(60-45)

     = \big(\cos(90)\cos(60)-\sin(90)\sin(60)\big)\big(\cos(60)\cos(45)+\sin(60  )\sin(45)\big) -  \big(\sin(90)\cos(60) + \sin(60)\cos(90)\big)\big(\sin(60)\cos(45)-\sin(45)\cos(60)\big)

    And you know that:

     \cos(90) = 0

     \sin(90) = 1

     \cos(60) = \frac{1}{2}

     \sin(60) = \frac{\sqrt{3}}{2}

     \cos(30) = \frac{\sqrt{3}}{2}

     \sin(30) = \frac{1}{2}

     \cos(45) = \frac{1}{\sqrt{2}}

     \sin(45) = \frac{1}{\sqrt{2}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    197
    Quote Originally Posted by Mush View Post
    Hrm not quite.

     \cos(165) = \cos(150+15) = \cos(150)\cos(15)-\sin(150)\sin(15)

     = \cos(90 + 60)\cos(60-45) - \sin(90+60) \sin(60-45)

     = \big(\cos(90)\cos(60)-\sin(90)\sin(60)\big)\big(\cos(60)\cos(45)+\sin(60  )\sin(45)\big) -  \big(\sin(90)\cos(60) + \sin(60)\cos(90)\big)\big(\sin(60)\cos(45)-\sin(45)\cos(60)\big)

    And you know that:

     \cos(90) = 0

     \sin(90) = 1

     \cos(60) = \frac{1}{2}

     \sin(60) = \frac{\sqrt{3}}{2}

     \cos(30) = \frac{\sqrt{3}}{2}

     \sin(30) = \frac{1}{2}

     \cos(45) = \frac{1}{\sqrt{2}}

     \sin(45) = \frac{1}{\sqrt{2}}
    Can you use the diagram that you gave me to solve this problem please so it can be extremely clear ?
    Thank you and sorry if i ask too much
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    777
    Hello, mj.alawami!

    Question: cos 165

    can it be solved by using the special angle
    cos 150 + cos 15
    cos 150 + cos (60-45)

    Can it be solved in this way and will it provide me the correct answer? . . . . Yes!
    It is simpler with only two angles, of course . . .


    Try: . \cos165^o \;=\;\cos(120^o + 45^o)

    . . . . . . . . . = \;\cos120^o\cos45^o - \sin120^o\sin45^o

    . . . . . . . . . = \;\left(-\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2  }}{2}\right)

    . . . . . . . . . = \;-\frac{\sqrt{2}}{4} - \frac{\sqrt{2}\sqrt{3}}{4}

    . . . . . . . . . =\;-\frac{\sqrt{2}(1 + \sqrt{3})}{4}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving Sum and difference identities
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 28th 2009, 09:44 AM
  2. [SOLVED] Sum and difference identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 25th 2009, 12:13 PM
  3. Sum and difference identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 9th 2009, 08:59 PM
  4. Help with sum and difference identities, please!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 7th 2008, 12:32 PM
  5. Sum Difference Identities
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 18th 2008, 02:31 PM

Search Tags


/mathhelpforum @mathhelpforum