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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?
-1
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
