3cos
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No, but cos(-x) = cos(x), because the cosine function is an even function.
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 x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
There is no reason at all. For most angles sin plus cos do not equal one.
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec