For any real number x such that 0 < x < 3, which of the following expressions is equivalen… — ACT

For any real number $x$ such that $0 < x < 3$ , which of the following expressions is equivalent to $\sin\left(\arccos\left(\frac{x}{3}\right)\right)$ ?

$\frac{\sqrt{9 - x^2}}{3}$ Cevap
B
$\frac{3 - x}{3}$
C
$\frac{\sqrt{x^2 - 9}}{3}$
D
$\frac{x}{\sqrt{9 - x^2}}$
E
$\frac{3}{x}$

Cevap

\frac{\sqrt{9 - x^2}}{3}

By letting

\theta = \arccos\left(\frac{x}{3}\right)

, we construct a right triangle with an adjacent side of length

x

and a hypotenuse of length

3

. Applying the Pythagorean theorem yields an opposite side length of

\sqrt{3^2 - x^2} = \sqrt{9 - x^2}

. The sine of

\theta

is the ratio of the opposite side to the hypotenuse, which evaluates to

\frac{\sqrt{9 - x^2}}{3}

Adım Adım Çözüm

Define an angle variable for the inverse trigonometric expression

Let

\theta = \arccos\left(\frac{x}{3}\right)

, so

\cos(\theta) = \frac{x}{3}

for

0 < \theta < \frac{\pi}{2}

Setting the inverse trigonometric function to an angle allows setting up a right triangle relationship.

Set up side lengths of a right triangle using the definition of cosine

Adjacent side =

x

, Hypotenuse =

3

By definition,

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}

Calculate the opposite side using the Pythagorean theorem

\text{opposite} = \sqrt{3^2 - x^2} = \sqrt{9 - x^2}.

In any right triangle,

\text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2

Determine the sine of the angle

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{9 - x^2}}{3}.

By definition,

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}

Anahtar Kavram

Evaluating algebraic compositions of trigonometric and inverse trigonometric functions using right triangle geometry.

Tahmini Süre:1m 15s

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