It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines.
sin x / (1 - cos x) = csc x + cot x
sin x / (1 - cos x) = 1 / sin x + cos x / sin x
Now it should be easy to do some simplifications:
sin x / (1 - cos x) = (1 + cos x) / sin x
Multiply both sides by 1 + cos x:
sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x
sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x
sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x
sin x (1 + cos x) / sin x = (1 + cos x)2
1 + cos x = (1 + cos x)2
1 = 1 + cos x
cos x = 0
So, cos x can be pi/2, 3 pi / 2, etc.
In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
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
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
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
1/sin x = csc x
The value of tan and sin is positive so you must search quadrant that tan and sin value is positive. The only quadrant fill that qualification is Quadrant 1.
[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,
The Answers.com interface does not handle symbols. Generally, algebraic expressions need to be written out with words like equals, plus, minus, multiplied by, divided by, etc.
sin7x-sin6x+sin5x
Sin 15 + cos 105 = -1.9045
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
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.
Sin(pi/6) = 1/2 is a true statement [not pie].
sex plus sin equals to lust
You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0
Assuming the question refers to [sin(x)]/2 rather than sin(x/2) the answer is 1.
No. But that could be because of limitations of the browser used by Answers.com. This means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals".
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