math
posted by mysterychicken on .
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

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

#1
sin2xcos2x = 4sin2x
sin2x(cos2x4) = 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 halfangle 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) (√(42√3)√(4+2√3))
= 1/2√2 ((√31)(√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. 
I asked my teacher for answer choices for #2 and she said root2/2, root2/2, or root3/4...
you said 1root2 
as you surely know by now
1/√2 = 1/√2 * √2/√2 = √2/2
c'mon, man. 
Gotcha!

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

Looks good to me.
Did you crank that out manually? I wen to
wolframalpha.com and typed in
√(s(sa)(sb)(sc)) where a=4.8 and b=6.3 and c=7.5 and s=(a+b+c)/2 
Yes I got the formula and plugged the values in :)