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URGENT PRE-CALCULUS HELP!?

+3 votes

sin α = 4/5, π/2 < α < π
cos β = 5/13, 0 < β < π
Find cos (α + β) , tan (α - β) , sin2α , and cos(β/2)

Please show work! This is the last part of an assignment I have and it's the only part I don't get!

asked Jan 16, 2013 in PRECALCULUS by skylar Apprentice

4 Answers

+2 votes

   sin α = 4/5, π/2 < α < π
   cos β = 5/13, 0 < β < π

     sin² α + cos² α =1

     (4/5)² +  cos² α =1

               cos² α = 1 - (16/25)

               cos² α =  9/25

               cos  α =  3/5

    sin² β + cos² β =1

    sin² β + (5/13)² = 1

               sin² β =  1 - 25/169

               sin² β =  144/169

               sin β =   12/13

1)  cos(α + β) = cosα cosβ - sinα sinβ

                         = (3/5)(5/13) - (4/5)(12/13)

                         =  15/65 - 48/65

                         =  -33/65

     cos(α + β)  =   -33/65

answered Jan 16, 2013 by Johncena Apprentice
+3 votes

2)  tan (α - β)

     tan α = sinα/ cosα

               = (4/5)/(3/5)

               = 4/3

    tan β  = sinβ/ cosβ

              =  (12 /13)/(5/13)

              =  12/5

   tan (α - β) = (tanα - tanβ)/(1- tanα tanβ)

                    = (4/3 - 12/5)/(1- (4/3)(12/5) )

                    =  (-16/15) / (1- 16/5)

                   =  (-16/15) / (-11/5)

                   =  16/33

     tan (α - β) =  16/33

answered Jan 16, 2013 by Johncena Apprentice
+5 votes

3)  sin2α

    sin2α =  2 sinα cosα

              =  2 (4/5) (3/5)

              =  24/25

   sin2α  = 24/25

answered Jan 16, 2013 by Johncena Apprentice
+3 votes

4) cos(β/2)

   cos (β/2) = ±√(1+ cos β)/2

                  = ± √(1+ 5/13)/2

                  = ± √(18/13)/2

                  = ± √18/26

  cos (β/2)   = ± √18/26

 

answered Jan 16, 2013 by Johncena Apprentice

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