If you mean:
sin2(x) cos2(x)
then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further:
sin(x)2cos(x)2
= [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2]
= (1 - cos(2x))(1 + cos(2x)) / 2
= (1 - cos2(2x)) / 2
Also note that 1 - cos2(x) = sin2(x), so we can then say:
= sin2(2x) / 2
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f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)