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advanced trigonometry?
For -pie/2<x<pie/2 give your answer in terms of pie
a) sin 2x = cos 2x
b) 5 sin x - 14= 2/sin x
1 Answer
- The Integral ∫Lv 79 years agoFavorite Answer
sin(2x) = cos(2x)
2sin(x)cos(x) = 1 - sin^2(x) ----> square both sides
4sin^2(x)cos^2(x) = ( 1 - sin^2(x) )^2
4sin^2(x)(1 - sin^2(x)) - ( 1 - sin^2(x) )^2 = 0
(1 - sin^2(x)) * [ 4sin^2(x) - ( 1 - sin^2(x) ) ] = 0
(1 - sin^2(x)) * [ 4sin^2(x) - 1 + sin^2(x) ] = 0
(1 - sin^2(x)) * (5sin^2(x) - 1) = 0
(1 - sin^2(x)) = 0
1 = sin^2(x)
+/- 1 = sin(x) . . . .sin(x) = 1 --> x = π/2
sin(x) = -1 --> x = -π/2
(5sin^2(x) - 1) = 0 ---> sin^2(x) = 1/5 ---> sin(x) = +/- 1/√(5)
sin(x) = 1/√(5)
x = sin^-1( 1/√(5) )
sin(x) = -1/√(5)
x = sin^-1( -1/√(5) )
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b)
5 sin(x - 14) = 2/sin(x) <--- do u mean this???