Best Answer

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

Q: How do you solve cot parenthesis sin to the negative 1 parenthesis 2 over 3 closed parenthesis and another closed parenthesis?

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Parenthesis

Your equation has two variables in it ... 'a' and 'x'. So the solution is a four-step process: 1). Get another independent equation that relates the same two variables. 2). Solve one of the equations for one of the variables. 3). Substitute that into the other equation, yielding an equation in a single variable. Solve that one for the single variable. 4). Substitute that value back into the first equation, and solve it for the second variable.

You expand the parenthesis out first. Then simplify the rest.

You are given: G(x) = x2 + x So if you want to solve G(2), all you need to do is replace all occurrences of "x" with the number 2, then work out what it comes to. G(2) = 22 + 2 ∴G(2) = 4 + 2 ∴G(2) = 6

Yes, and parenthesis are the first thing you solve.

Related questions

sin(arcsin(2/3)) = 2/3, since sin is the inverse function of arcsin.

tan(sec-1(5/2))Start with sec-1(5/2), which is the same as cos-1(2/5). So there is a right triangle, where the side adjacent the angle is 2, and the hypotenuse is 5. Solve for the opposite side: sqrt(5Â² - 2Â²) = sqrt(21).Tangent is opposite over adjacent, so the answer is sqrt(21)/2

2

Parenthesis

You expand the parenthesis out first. Then simplify the rest.

Your equation has two variables in it ... 'a' and 'x'. So the solution is a four-step process: 1). Get another independent equation that relates the same two variables. 2). Solve one of the equations for one of the variables. 3). Substitute that into the other equation, yielding an equation in a single variable. Solve that one for the single variable. 4). Substitute that value back into the first equation, and solve it for the second variable.

You are given: G(x) = x2 + x So if you want to solve G(2), all you need to do is replace all occurrences of "x" with the number 2, then work out what it comes to. G(2) = 22 + 2 ∴G(2) = 4 + 2 ∴G(2) = 6

Yes, and parenthesis are the first thing you solve.

Let y = sin(cos-1(2/5)) Suppose x = cos-1(2/5): that is, cos(x) = 2/5 then sin2(x) = 1 - cos2(x) = 1 - 4/25 = 21/25 so that sin(x) = sqrt(21)/5 which gives x = sin-1[sqrt(21)/5] Then y = sin(cos-1(2/5)) = sin(x) : since x = cos-1(2/5) =sin{sin-1[sqrt(21)/5]} = sqrt(21)/5 There will be other solutions that are cyclically related to this one but no range has been given for the solutions.

Parentheses is when you are doing an equation, and you solve the problem.

7a=2(-10)........................................answer is 20/7

You square the number in the parentheses.