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Limit x approaches 0 sinx divide x equals 1 evaluate limit x approaches 0 cos2x-1 divide x?
== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.
Find all radian solutions in the interval 0 is less than or equal to x less than 2pi 2sin2 x - 6sinx - 1 equals 0?
2sin2x - 6sinx - 1 = 0Therefore, using the quadratic equation,sinx = (3-sqrt(11)/2 = -0.1583 or sinx > 3.The latter solution is not possible since |sin(x)| cannot exceed 1.arcsin(-0.1583) = -0.1590 radiansso x = 2pi - 0.1590 = 6.1242 radiansalso x = pi + 0.1590 = 3.3006 radians
How do you find the solution for 2sin2 plus cos2 equals 1?
Assume the expression is: 2sin²(x) + cos²(x) = 1 Use this identity to work out the problem: cos²(x) = 1 - sin²(x) Then: 2sin²(x) + 1 - sin²(x) = 1 sin²(x) + 1 = 1 sin²(x) = 0 sin(x) = 0 Therefore, the solutions are: x = πk radians where k belongs to the set of integers In degrees, we obtain: x = 180l° where l belongs to the set of integers
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
What is 2sin2 in radians?
1.8185948536513634
Limit x approaches 0 sinx divide x equals 1 evaluate limit x approaches 0 cos2x-1 divide x?
== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.
Find all radian solutions in the interval 0 is less than or equal to x less than 2pi 2sin2 x - 6sinx - 1 equals 0?
2sin2x - 6sinx - 1 = 0Therefore, using the quadratic equation,sinx = (3-sqrt(11)/2 = -0.1583 or sinx > 3.The latter solution is not possible since |sin(x)| cannot exceed 1.arcsin(-0.1583) = -0.1590 radiansso x = 2pi - 0.1590 = 6.1242 radiansalso x = pi + 0.1590 = 3.3006 radians
How do you find the solution for 2sin2 plus cos2 equals 1?
Assume the expression is: 2sin²(x) + cos²(x) = 1 Use this identity to work out the problem: cos²(x) = 1 - sin²(x) Then: 2sin²(x) + 1 - sin²(x) = 1 sin²(x) + 1 = 1 sin²(x) = 0 sin(x) = 0 Therefore, the solutions are: x = πk radians where k belongs to the set of integers In degrees, we obtain: x = 180l° where l belongs to the set of integers
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What does Sin squared x - Cos squared x equal?
1 - 2cos2(x) and also 2sin2(x) - 1 Take your choice. Use whichever one is more convenient.
What is sine theta plus 2 cosine squared theta?
I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
What is the second derivative of y equals 2sinx cosx?
y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
What is the work for 2 sin squared x equals 2 plus cos x?
2sin2 x2(1 - cos2 x) = 2 + cos x2 - 2cos2 x = 2 + cos x- 2cos2 x - cos x = 0cos x(-cos x - 1) = 0cos x = 0 or -cos x - 1 = 0cos x = -1Since the cosine has a period of 2pi, in the interval 0 ≤ x ≤ 2piwhen cos x = 0, x = pi/2 and x = 3pi/2, andwhen cos x = -1, x = pi.All values of x that satisfy the given equation are:x = pi/2 + 2npi, x = pi +2npi, and x = 3pi/2+ 2npi where n is any integer.
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
