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.
Often is another word for over and over again. Additional synonyms include repeatedly and repetitious.
leap over
Another word for " over"
Again , repeatedly, a lot, another time, unending, infinite, often, continuously.
atop
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(5² - 2²) = sqrt(21).Tangent is opposite over adjacent, so the answer is sqrt(21)/2
one over 3 a3
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.
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
13/21 - 7/21 = 6/21 Sorry, DJ, 4 over 7 = 12/21 so answer is 5/21
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.
(DCCL)I or instead of the parenthesis, you can use a line over the top, but I can't write that on here. Both parenthesis and a line above indicate x1000
You could try putting that part in parenthesis. In algebra, formula parts in parenthesis are worked separately and take precedence over what is no in parenthesis. Things are usually worked left to right, and multiplication or division before addition or subtraction.
No, if the vent is closed it should be no problem.
The distributive property is the ability of one operation to "distribute" over another operation contained inside a set of parenthesis. Most commonly, this refers to the property of multiplication distributing over addition or subtraction, such that x(a+b) = xa + xb. When we say that multiplication distributes over addition, it means we can distribute the factor outside the set of parenthesis to each item inside, and then add the results. For example, 4(3+7) is equivalent to 4*3 + 4*7 because the multiplication by four was distributed across the addition inside the parenthesis. Not every operation is distributive. For example, division is not distributive over addition. If we are given 20/(3+7) the true result is 2, but distributing would give you 20/3 + 20/7, which is around 10 and very incorrect!
I will assume it is written out as follows with no parenthesis... 3 / 3 / 5 I would stick to the order of operations, and go straight through it... 3 / 3 / 5 = 1 / 5 (or one-fifth) If there are parenthesis involved, they take priority.