Take a right angled triangle ABC with the right angle at B, so that AC is the hypotenuse.
Let AC be 1 unit long.
Using the angle CAB, the length of AC and the trigonometric ratios:
sin = opposite/hypotenuse ⇒ sin CAB = AB/AC = AB/1 = AB
cos = adjacent/hypotenuse ⇒ cos CAB = BC/AC = BC/1 = BC
Using Pythagoras:
AB2 + BC2 = AC2
⇒ (sin CAB)2 + (cos CAB)2 = 12
⇒ sin2θ + cos2θ = 1
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cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
Cosec = -9 => sin = -1/9 Then, cos2 = 1 - sin2 = 1 - (-1/9)2 = 1 - 1/81 = 80/81 and so cos = sqrt(80/81) = ±8.94/9 = ±0.99