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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

[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,

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

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.

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°

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

sex plus sin equals to lust

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec

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))

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.

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

Christianity provides a path to forgiveness with sin; it tries to encourage people to avoid sin.

The differentiation of sin x plus cosx is cos (x)-sin(x).

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.

leonhard euler

Angle A = 52° 15' = 52 25° therefore angle B = 90 - 52.25 = 37.75°. Using the Sine Rule : a/sin A = b/sin B. 6.7808/sin 52.25 = b/sin 37.75 : b = 6.7808 sin 37.75 ÷ sin 52.25 = 5.2503 Either using the Sine Rule or Pythagoras gives the length of the hypotenuse as 8.5758

Do sin(x), square it, and then multiply it by two.

Assuming sin equals 0.3237, the angle is in quadrant I.

28 The Law of Sines: a/sin A = b/sin B = c/sin C 24/sin 42˚ = c/sin (180˚ - 42˚ - 87˚) since there are 180˚ in a triangle. 24/sin 42˚ = c/sin 51˚ c = 24(sin 51˚)/sin 42˚ ≈ 28

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

The differentiation of sin x plus cosx, yields cos(x)-sin(x).

22.20366435 sin^-1(0.3779)

2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210Â°, 330Â°or sin(x) - 1 = 0sin(x) = 1x = 90Â°

sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.