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sin is short for sine. Sin(x) means the ratio of the side of a right triange opposite the angle 'x' divided by the length of the hypotenuse.
cos is short for cosine. Cos(x) is equal to the similar ratio of the side adjacent to the angle 'x' divided by the length of the hypotenuse.
tan is short for tangent. Tan(x) is equal to the ratio of the opposite side divided by the adjacent side. This is the same as sin(x)/cos(x).
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How tan9-tan27-tan63 tan81 equals 4?
tan(9) + tan(81) = sin(9)/cos(9) + sin(81)/cos(81)= {sin(9)*cos(81) + sin(81)*cos(9)} / {cos(9)*cos(81)} = 1/2*{sin(-72) + sin(90)} + 1/2*{sin(72) + sin(90)} / 1/2*{cos(-72) + cos(90)} = 1/2*{sin(-72) + 1 + sin(72) + 1} / 1/2*{cos(-72) + 0} = 2/cos(72) since sin(-72) = -sin(72), and cos(-72) = cos(72) . . . . . (A) Also tan(27) + tan(63) = sin(27)/cos(27) + sin(63)/cos(63) = {sin(27)*cos(63) + sin(63)*cos(27)} / {cos(27)*cos(63)} = 1/2*{sin(-36) + sin(90)} + 1/2*{sin(72) + sin(36)} / 1/2*{cos(-36) + cos(90)} = 1/2*{sin(-36) + 1 + sin(36) + 1} / 1/2*{cos(-36) + 0} = 2/cos(36) since sin(-36) = -sin(36), and cos(-36) = cos(36) . . . . . (B) Therefore, by (A) and (B), tan(9) - tan(27) - tan(63) + tan(81) = tan(9) + tan(81) - tan(27) - tan(63) = 2/cos(72) – 2/cos(36) = 2*{cos(36) – cos(72)} / {cos(72)*cos(36)} = 2*2*sin(54)*sin(18)/{cos(72)*cos(36)} . . . . . . . (C) But cos(72) = sin(90-72) = sin(18) so that sin(18)/cos(72) = 1 and cos(36) = sin(90-36) = sin(54) so that sin(54)/cos(36) = 1 and therefore from C, tan(9) – tan(27) – tan(63) + tan(81) = 2*2*1*1 = 4
What are the double-angle and half-angle identities?
sin 2θ = 2(sin θ)(cos θ) cos 2θ = (cos θ)2 - (sin θ)2 cos 2θ = 2(cos θ)2 - 1 cos 2θ = 1 - 2(sin θ)2 tan 2θ = 2(tan θ)/[1 - (tan θ)2] sin θ/2 = ±√[(1 - (cos θ))/2] cos θ/2 = ±√[(1 + (cos θ))/2] tan θ/2 = ±√[(1 - (cos θ))/(1 + (cos θ))] ; cos θ ≠ -1 tan θ/2 = [1 - (cos θ)]/(sin θ) tan θ/2 = (sin θ)/[1 + (cos θ)]
What is the solution for cos tan csc equals 1?
Well I don't exactly get "the solution", but simplifying the equation is quite simple. Maybe that's what you're looking for. Here are the steps for simplifying it. costancsc = 1 1. Change tan to sin/cos 2. Change csc to 1/sin cos(sin/cos)(1/sin) = 1 And as you can now see, the first cos cancels with the second one under the sin/cos fraction, and the first sin cancels with the second one under the 1/sin fraction consequentially leaving you with 1 = 1. For a better look, notice this fraction when all three parts are combined cos * sin * 1 ---------------- cos * sin See how the cos and sin cancel each other leaving you with 1 * 1 * 1 which is just 1. Therefore the final simplification is just 1 = 1. I hope this helps!
How do you show that sinxcosxtanx equals 1-cos2x?
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How do you express cosine in terms of cotangent?
cos(x)=sin(x-tau/4) tan(x)=sin(x)/cos(x) sin(x)=tan(x)*cos(x) cos(x)=tan(x-tau/4)*cos(x-tau/4) you can see that we have some circular reasoning going on, so the best we can do is express it as a combination of sines and cotangents: cos(x)=1/cot(x-tau/4)*sin(x-tau/2) tau=2*pi
