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Does csc -120 equal -csc 120?
-240
Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?
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)
How do you solve csc x sin x equals cos x cot x plus?
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)
How do you find the derivative of - csc x - sin x?
d/dx (-cscx-sinx)=cscxcotx-cosx
What is the derivative of tan x?
The derivative of tan(x) is sec2(x).(Which is the same as 1/cos2(x).
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
Csc x tan x?
If you want to simplify that, it usually helps to express all the trigonometric functions in terms of sines and cosines.
What is the derivative of 3tanx-4cscx?
7
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
How do you make 'cot' and 'csc' on a TI-84 graphing calculator?
From math class, some trigonometric identities: cot x = 1/tan x csc x = 1/sin x sec x = 1/cos x There are no built-in cot or csc formulas, so use the above. Remember that these give errors when tan x, sin x, or cos x are equal to 0.
How do you simplify csc theta cot theta?
There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
What is sin cos tan csc sec cot of 135 degrees?
All those can be calculated quickly with your calculator. Just be sure it is in "degrees" mode (not in radians). Also, use the following identities: csc(x) = 1 / sin(x) sec(x) = 1 / cos(x) cot(x) = 1 / tan(x) or the equivalent cos(x) / sin(x)
Csc squared divided by cot equals csc x sec. can someone make them equal?
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
What is the second derivative of ln(tan(x))?
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2
Are shifts sin cos tan equal to csc sec cot respectively?
No, they are the inverse functions, while csc, sec and cot are the reciprocal functions. To illustrate the difference, the inverse of f(x) = x+3 is f-1(x) = x-3 But the reciprocal of f(x) is 1/f(x) = 1/(x+3)
