Q: What is Cos 15?

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Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)

Sin 15 + cos 105 = -1.9045

cos(15 deg) = 0.9659, approx.

The rate at which the lit portion of the moon moves per hour changes with latitude. The formula for finding the average rate of rotation per hour is: 15°cos(latitude). At the equator the equation would be 15°cos(0°)= 15° per hour.

Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.

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Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)

cos(195) = cos(180 + 15) = cos(180)*cos(15) - sin(180)*sin(15) = -1*cos(15) - 0*sim(15) = -cos(15) = -cos(60 - 45) = -[cos(60)*cos(45) + sin(60)*sin(45)] = -(1/2)*sqrt(2)/2 - sqrt(3)/2*sqrt(2)/2 = - 1/4*sqrt(2)*(1 + sqrt3) or -1/4*[sqrt(2) + sqrt(6)]

Sin 15 + cos 105 = -1.9045

It is: cos(15) = (sq rt of 6+sq rt of 2)/4

cos(15 deg) = 0.9659, approx.

There is some kind of formula here, half angle, or some such that I forget, but I do remember the algorithm. So...,int[cos(10X)cos(15X)] dxsince this is multiplicative, switch it aroundint[cos(15X)cos(10X)] dxint[cos(15X - 10X)/2(15 -10) + cos(15X + 10X)/2(15 + 10)] dxint[cos(5X)/10 + cos(25X)/50] dx= 1/10sin(5X) + 1/50sin(25X) + C=========================

csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4

The rate at which the lit portion of the moon moves per hour changes with latitude. The formula for finding the average rate of rotation per hour is: 15°cos(latitude). At the equator the equation would be 15°cos(0°)= 15° per hour.

4, 6 and 15 or 46 and 15 cos 4,6 and 15 LCM is 360 (4 x 6 x 15)

look at each other and have a friendly chat about the weather

Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.

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