# What algebraic expression is equivalent to the expression cos(arcsin((a-x)/c))?

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The algebraic expression that you want simplified is cos(arc sin((a-x)/c)). You should make proper use of brackets to represent your expression accurately.

cos(arc sin((a-x)/c))

=> sqrt [1 - (sin(arc sin((a-x)/c)))^2]

=> sqrt[1 - ((a-x)/c))^2]

=> sqrt[ c^2 - (a - x)^2]/c

=> sqrt[c^2 - a^2 - x^2 + 2ax]/c

**The required equivalent expression is sqrt[c^2 - a^2 - x^2 + 2ax]/c**

It is important to remember that arcsin ((a-x)/c) means the angle whose sin is ((a-x)/c).

To understand this, draw a right triangle.

Label the hypotenuse c.

Select one of the acute angles( call it n) as your angle and label the opposite side as (a-x).

Label the third side y.

Find the third side by using the pythagorean theorem:

c^2 - (a-x)^2 =y^2

expanding this formula results in:

c^2 -(a^2 -2ax + x^2) = y^2

c^2 -a^2 +2ax -x^2 = y^2.

Therefore y = sqrt(c^2 -a^2 +2ax -x^2)

and the cos of your angle is y/c or sqrt(c^2 -a^2 -x^2 +2ax)/c.

Let arcsin(a-x)/c = y, therefore sin [arcsin(a-x)/c] = sin y = (a-x)/c

Using Pythagorean identity, we'll get cos t:

cos y = sqrt[1 - (sin y)^2]

cos y = sqrt[1 - (a-x)^2/c^2]

cos y = sqrt {[c^2 - (a-x)^2]/c^2}

The difference of sqares returns the product:

cos y = sqrt [(c - a + x)(c + a - x)/c^2]

cos y = [sqrt (c - a + x)(c + a - x)]/c

**The algebraic expression equivalent to cos[arcsin(a-x)/c] is [sqrt (c - a + x)(c + a - x)]/c.**