How to find negative cosine

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ACT Math › How to find negative cosine

Questions 1 - 5

If $cos(x)=-\frac{\sqrt{3}}{2}$ and $0 \le x \le \pi$ , what is the value of $cos(x+\pi)$ ?

$\frac{1}{2}$

$\frac{2}{\sqrt{3}}$

$2\sqrt{5}$

$\frac{\sqrt{3}}{3}$

Explanation

Based on this data, we can make a little triangle that looks like:

Rt1

This is because $cos(x)=\frac{adjacent}{hypotenuse}$ .

Now, this means that must equal $\frac{\pi}{2}+\frac{\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}$ . (Recall that the cosine function is negative in the second quadrant.) Now, we are looking for:

$cos(\frac{2\pi}{3}+\pi)$ or $cos(\frac{5\pi}{3})$ . This is the cosine of a reference angle of:

$2\pi-\frac{5\pi}{3}=\frac{\pi}{3}$

Looking at our little triangle above, we can see that the cosine of $\frac{\pi}{3}$ is $\frac{1}{2}$ .

What is the cosine of the angle formed between the origin and the point if that angle is formed with one side of the angle beginning on the -axis and then rotating counter-clockwise to ? Round to the nearest hundredth.

Explanation

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.") Now, it is easiest to think of this like you are drawing a little triangle in the second quadrant of the Cartesian plane. It would look like:

Cos37

So, you first need to calculate the hypotenuse:

$h=\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{56}$

So, the cosine of an angle is:

$\frac{adjacent}{hypotenuse}$ or, for your data, $\frac{3}{\sqrt{56}}$ .

This is approximately . Rounding, this is . However, since is in the second quadrant your value must be negative: .

On a grid, what is the cosine of the angle formed between a line from the origin to and the x-axis?

$-\frac{\sqrt{10}}{10}$

$\frac{\sqrt{10}}{10}$

$\frac{3\sqrt{10}}{10}$

$-\frac{3\sqrt{10}}{10}$

$\textup{None of the answers provided is correct.}$

Explanation

If the point to be reached is , then we may envision a right triangle with sides and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that: .

Thus, $c = \sqrt{90} = 3\sqrt{10}$ .

Now, SOHCAHTOA tells us that $\textup{cosine} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , so we know that:

$\textup{cos} x^{\circ} = \frac{3}{3\sqrt{10}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$

Thus, our cosine is approximately $\frac{\sqrt{10}}{10}$ . However, as we are in the second quadrant, cosine must be negative! Therefore, our true cosine is $-\frac{\sqrt{10}}{10}$ .

To the nearest , what is the cosine of the angle formed between the origin and ? Assume a counterclockwise rotation.

Explanation

If the point to be reached is , then we may envision a right triangle with sides and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that:

Thus, $c = \sqrt{212} = 2\sqrt{53}$

Now, SOHCAHTOA tells us that $\textup{cosine}= \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , so we know that:

$cos x^{\circ} = \frac{4}{2\sqrt{53}} \approx 0.2747$

Thus, our cosine is approximately . However, as we are in the third quadrant, cosine must be negative! Therefore, our true cosine is .

Explanation

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like:

Cos611

So, you first need to calculate the hypotenuse. You can do this by using the Pythagorean Theorem, , where and are the lengths of the legs of the triangle and the length of the hypotenuse. Rearranging the equation to solve for , you get: