cos60= 1/2
sin60=1.732/2
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(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
Cos (x) = -Sin(x) 1 = -Sin(x) / Cos (x) 1 = -Tan(x) Tan(x) = -1 x = Tan^-1(-1( x = -45 degrees = - pi /4 , 3pi/4, 5pi/4 ....
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞