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What is x if the sin of x equals square root of 5 divided by 2?
2.5
How do you solve cot parenthesis sin to the negative 1 parenthesis 2 over 3 closed parenthesis and another closed parenthesis?
I presume that sin-1x is being used to represent the inverse sin function (I prefer arcsin x to avoid possible confusion). Make use of the trignometirc relationships: cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - sin2θ)/sinθ sin(arcsin x) = x Then: cot(arcsin(x)) = √(1 - sin2(arcsin(x))/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(2/3)) = √(1 - (2/3)2)/(2/3) = √(9/32 - 4/32) ÷ 2/3 = √(9 - 4) x 1/3 x 3/2 = 1/2 x √5
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
What expression is undefined when y equals 4?
There are infintely many posibilities: Any expression that contains (y-4) or a multiple of it in the denominator. Also, the following trigonometric functions: An expression that contains arcsin(x/y) where abs(x) < 4 Similarly arccos Corresponding conditions for arccosec and arcsec.
How do you solve 2 sin squared x minus sinx minus 1 is equal to 0?
2 sin(x)2 - sin(x) - 1 = 0 Let Y=sin(x) then the equation is 2*Y2 - Y - 1 =0 Delta = (-1 * -1) - 4 * 2 * -1 = 9 Y = (1 + sqrt(9)) / 4 or Y = (1 - sqrt(9)) / 4 Y = 1 or Y = -1/2 Then x = Arcsin(Y) and (in radians) x = Arcsin(1) = Pi/2 +2*k*Pi or x=Arcsin(-1/2) = -Pi/6 + 2*k*Pi where k is an integer
What is the difference between Arcsin x and arcsin x?
They're the same. They're the same.
What is the integral of arcsinxdx?
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
How do you solve cot parenthesis inverse of sin 4 over 7 closed parenthesis?
The inverse sin function I write as arcsin x. Make use of the trignometric relationships: cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - (sinθ)2)/sinθ sin(arcsin x) = x Then: cot(arcsin(x)) = √(1 - (sin(arcsin(x))2)/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(4/7)) = √(1 - (4/7)2)/(4/7) = √(49/72 - 16/72) ÷ 4/7 = √(49 - 16) x 1/7 x 7/4 = 1/4 x √33
What is x if the sin of x equals square root of 5 divided by 2?
2.5
How do you solve cot parenthesis sin to the negative 1 parenthesis 2 over 3 closed parenthesis and another closed parenthesis?
I presume that sin-1x is being used to represent the inverse sin function (I prefer arcsin x to avoid possible confusion). Make use of the trignometirc relationships: cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ/sinθ = √(1 - sin2θ)/sinθ sin(arcsin x) = x Then: cot(arcsin(x)) = √(1 - sin2(arcsin(x))/sin(arcsin(x)) = √(1 - x2)/x ⇒ cot(arcsin(2/3)) = √(1 - (2/3)2)/(2/3) = √(9/32 - 4/32) ÷ 2/3 = √(9 - 4) x 1/3 x 3/2 = 1/2 x √5
Does the notation of arcsin x represent the inverse function to sine?
NO FALSE
What is cot parenthesis sin to the negativde 1 parenthesis 2 over 3 closed parenthesis and another closed parenthesis?
If I read that correctly, you have: cot(sin-1(2/3)) which I understand to mean cot(arcsin(2/3)) which has the value 1/2 x √5 sin(arcsin(x)) = x cos2θ + sin2θ = 1 ⇒ cosθ = √(1 - sin2θ) cotθ = cosθ ÷ sinθ ⇒ cot(arcsin(2/3)) = cos(arcsin(2/3)) ÷ sin(arcsin(2/3) = √(1 - sin2(arcsin(2/3))) ÷ sin(arcsin(2/3) = √(1 - (2/3)2) ÷ (2/3) = 1/3 x √(9 - 4) x 3/2 = 1/2 x √5 As the reciprocal trignometric functions have separate names, eg 1/tan x = cot x, the use of the -1 "power" to indicate the inverse function is possible. However, to avoid any possible confusion, I prefer to use the arc- prefix to indicate the inverse function.
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
What is x if the Sin of x equals square root 3 divided by 2?
1.5
What expression is undefined when y equals 4?
There are infintely many posibilities: Any expression that contains (y-4) or a multiple of it in the denominator. Also, the following trigonometric functions: An expression that contains arcsin(x/y) where abs(x) < 4 Similarly arccos Corresponding conditions for arccosec and arcsec.
What is the answer to 7 sin x divided by cosxsquared minus 6 cos x divided by sin xsquared equals 0?
(7sin(x))/(cos2(x)-6cos(x))/(sin2(x))=0(7sin3(x))/(cos2(x)-6cos(x))=07sin3(x)=0sin(x)=0x=arcsin(0)x=(pi)(n)where n is any whole numberWe only set the top part of the function equal to zero because we know that if the bottom function equals zero then the answer is undefined. Therefore the top function is the only one that will produce a value of zero.
How do you solve 2 sin squared x minus sinx minus 1 is equal to 0?
2 sin(x)2 - sin(x) - 1 = 0 Let Y=sin(x) then the equation is 2*Y2 - Y - 1 =0 Delta = (-1 * -1) - 4 * 2 * -1 = 9 Y = (1 + sqrt(9)) / 4 or Y = (1 - sqrt(9)) / 4 Y = 1 or Y = -1/2 Then x = Arcsin(Y) and (in radians) x = Arcsin(1) = Pi/2 +2*k*Pi or x=Arcsin(-1/2) = -Pi/6 + 2*k*Pi where k is an integer
