Solving for Cosine Given Sine in a Right Triangle

We are given a right triangle and an acute angle, $x$. We are told that the value of $\sin x$ is $\frac{3}{7}$. Our goal is to find the value of $\cos x$ for this acute angle $x$.

In trigonometry, for any acute angle $x$ in a right triangle, the sine and cosine values are related by a fundamental identity known as the Pythagorean identity. This identity states:

$$ \sin^2 x + \cos^2 x = 1 $$

Since we know the value of $\sin x$, we can substitute it into this identity to find the value of $\cos x$. We are given $\sin x = \frac{3}{7}$.

Substitute the given value into the identity:

$$ \left(\frac{3}{7}\right)^2 + \cos^2 x = 1 $$

Now, calculate the square of $\sin x$:

$$ \frac{9}{49} + \cos^2 x = 1 $$

To find $\cos^2 x$, subtract $\frac{9}{49}$ from both sides of the equation:

$$ \cos^2 x = 1 - \frac{9}{49} $$

To perform the subtraction, find a common denominator, which is 49:

$$ \cos^2 x = \frac{49}{49} - \frac{9}{49} $$

$$ \cos^2 x = \frac{49 - 9}{49} $$

$$ \cos^2 x = \frac{40}{49} $$

Now we have the value of $\cos^2 x$. To find $\cos x$, we need to take the square root of both sides. Since $x$ is an acute angle in a right triangle, the cosine value must be positive.

$$ \cos x = \sqrt{\frac{40}{49}} $$

We can simplify the square root by splitting the numerator and the denominator:

$$ \cos x = \frac{\sqrt{40}}{\sqrt{49}} $$

The square root of 49 is 7. For $\sqrt{40}$, we can simplify it by factoring out perfect squares. $40 = 4 \times 10$, and 4 is a perfect square ($\sqrt{4} = 2$).

$$ \sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10} $$

Substitute these simplified values back into the expression for $\cos x$:

$$ \cos x = \frac{2\sqrt{10}}{7} $$

So, the value of $\cos x$ for the given acute angle is $\frac{2\sqrt{10}}{7}$.

Revision Table: Finding Cosine

Additional Information on Trigonometric Ratios

In a right triangle, the trigonometric ratios (sine, cosine, tangent, etc.) relate the angles to the lengths of the sides. For an acute angle $x$:

• Sine ($ \sin x $) is the ratio of the length of the side opposite the angle to the length of the hypotenuse.
• Cosine ($ \cos x $) is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.
• Tangent ($ \tan x $) is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle ($ \tan x = \frac{\sin x}{\cos x} $).

The Pythagorean identity $ \sin^2 x + \cos^2 x = 1 $ is derived directly from the Pythagorean theorem ($ a^2 + b^2 = c^2 $) applied to the sides of a right triangle and dividing by the square of the hypotenuse.

For acute angles (angles between 0 and 90 degrees), both $\sin x$ and $\cos x$ are positive values.