90+ whatever number is in form of sin.
Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.
Every angle has a sine and a cosine. The sine of 35 degrees is 0.57358 (rounded) The cosine of 35 degrees is 0.81915 (rounded)
Because it is not always convenient to convert to convert sin to cos when working with adjacent sides.
it is the same as a sin function only shifted to the left pi/2 units
No.-1
cos(x) = sin(pi/2-x) = -sin(x-pi/2)
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
sin = opp/hyp cos = adj/hyp tan = opp/adj
The expression that completes the identity ( \sin u \cos v ) is ( \frac{1}{2} (\sin(u + v) - \sin(u - v)) ). This identity is derived from the product-to-sum formulas in trigonometry, which relate products of sine and cosine functions to sums and differences of sine functions.
cos = sqrt(1 - sin^2)
No, it does not.
the only close answer i know is: eix = cos(x)+i*sin(x) where i is imaginary unit
cos(35)sin(55)+sin(35)cos(55) If we rewrite this switching the first and second terms we get: sin(35)cos(55)+cos(35)sin(55) which is a more common form of the sin sum and difference formulas. Thus this is equal to sin(90) and sin(90)=1
f(x) = Cos(x) f'(x) = -Sin(x) Conversely f(x) = Sin(x) f'(x) = Cos(x) NB Note the change of signs.
Generally, the derivative of sine is cosine.
sin 0 = 0 cos 0 = 1
csc(x) = 1/sin(x) = +/- 1/sqrt(1-cos^2(x))