2 sin2 x + sin x = 1.
Letting s = sin x, we have:
2s2 + s - 1 = (2s - 1)(s + 1) = 0; whence,
sin x = ½ or -1, and
x = 30° or 150° or 270°.
Or, if you prefer,
x = π/6 or 5π/6 or 3π/2.
sin7x-sin6x+sin5x
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Do sin(x), square it, and then multiply it by two.
No, (sinx)^2 + (cosx)^2=1 is though
1
sin7x-sin6x+sin5x
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Do sin(x), square it, and then multiply it by two.
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
22
No, (sinx)^2 + (cosx)^2=1 is though
1
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
2 x cosine squared x -1 which also equals cos (2x)