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Can someone please check my answers and help me with the last question!

1. Solve sin2xcos2x = 4sin2x on the interval [0, 2pi]
0, pi, 2pi?

2. Exact value of sin(pi/12) - sin(5pi/12)
root3/4?

3. Using factorial notation, 0! = 1
False?

4. Find the area of the following triangle. a=4.8inches, b=6.3inches, c=7.5inches. Round answer to nearest tenth

  • math -

    area is 1/2bh...so can I make any side the base/height?

  • math -

    #1
    sin2xcos2x = 4sin2x
    sin2x(cos2x-4) = 0
    sin2x = 0 or cos2x = 4
    so, 2x = 0,π,2π,3π,4π
    x = 0,π/2,π,3π/2,2π
    naturally, cos2x=4 has no solutions.

    Again, you found solutions for 2x in [0,2π], but you want all values for x in [0,2π]

    #2 Since sin u − sin v = 2 sin(½(u−v)) cos(½(u+v)),

    sin(pi/12) - sin(5pi/12)
    Using the half-angle formula,
    = √((1 - cos pi/6)/2)-√((1 + cos pi/6)/2)
    = 1/√2 (√(1-√3/2)-√(1+√3/2))
    = 1/√2 (√((2-√3)/2) - √((2+√3)/2))
    = 1/√2 (1/2) (√(4-2√3)-√(4+2√3))
    = 1/2√2 ((√3-1)-(√3+1))
    = 1/2√2 (-2)
    = -1/√2

    Or, using the sum/difference formulas,
    sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))

    sin(pi/12) - sin(5pi/12)
    2sin(-pi/6)cos(pi/4)
    = 2(-1/2)(1/√2)
    = -1/√2

    #3 True. It is defined that way.

    4. yes, any side can be the base.
    so how do you get the height?
    You might consider Heron's formula.

  • math -

    I asked my teacher for answer choices for #2 and she said -root2/2, root2/2, or -root3/4...
    you said -1root2

  • math -

    as you surely know by now

    1/√2 = 1/√2 * √2/√2 = √2/2

    c'mon, man.

  • math -

    Gotcha!

  • math -

    I get 15.03 for the area...is that correct?

  • math -

    Looks good to me.
    Did you crank that out manually? I wen to

    wolframalpha.com and typed in

    √(s(s-a)(s-b)(s-c)) where a=4.8 and b=6.3 and c=7.5 and s=(a+b+c)/2

  • math -

    Yes I got the formula and plugged the values in :)

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