Q: How do you solve Sin x plus sin 3x plus sin 5x plus sin 7x equals?

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Sin 15 + cos 105 = -1.9045

y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5

Assuming the angles are expressed in radians:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + π∴ x = π/4On the other hand, if your angles are in degrees, then the answer would be:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + 180∴ x = 180°/4∴ x = 45°

The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.

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.

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Sin 15 + cos 105 = -1.9045

y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,

Assuming the angles are expressed in radians:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + π∴ x = π/4On the other hand, if your angles are in degrees, then the answer would be:sin(5x) + sin(x) = 0∴ sin(5x) = -sin(x)∴ 5x = x + 180∴ x = 180°/4∴ x = 45°

The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.

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

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.

Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)

2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))

sex plus sin equals to lust

No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1

2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.