Math - Understanding Sine, Cosine, and Tangent

What is the cosine of $30^\circ$ ?

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

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

$\frac{1}{2}$

$\frac{1}{3}$

$\sqrt{3}$

Explanation

The pattern for the side of a triangle is $x:x\sqrt{3}:2x$ .

Since $\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$ , we can plug in our given values.

$\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\cos=\frac{x\sqrt{3}}{2x}$

Notice that the 's cancel out.

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

Triangle

In the right triangle above, which of the following expressions gives the length of y?

$x\cos\theta$

$x\sin\theta$

$x\tan\theta$

$\cos(x\times \theta)$

$\frac{\sin(x)}{\theta}$

Explanation

$\cos\theta$ is defined as the ratio of the adjacent side to the hypotenuse, or in this case $\frac{y}{x}$ . Solving for y gives the correct expression.

If the polar coordinates of a point are $\left (2,\frac{-\pi}{4} \right )$ , then what are its rectangular coordinates?

$(\sqrt2,-\sqrt2)$

$(-\sqrt2,-\sqrt2)$

$(\sqrt2,\sqrt2)$

$(2,-\sqrt2)$

Explanation

The polar coordinates of a point are given as $(r,\theta )$ , where r represents the distance from the point to the origin, and $\theta$ represents the angle of rotation. (A negative angle of rotation denotes a clockwise rotation, while a positive angle denotes a counterclockwise rotation.)

The following formulas are used for conversion from polar coordinates to rectangular (x, y) coordinates.

$x=r\cos\theta$

$y=r\sin\theta$

In this problem, the polar coordinates of the point are $\left (2,\frac{-\pi}{4} \right )$ , which means that and $\theta=-\frac{\pi}{4}$ . We can apply the conversion formulas to find the values of x and y.

$x=r\cos\theta=2\cos\left (\frac{-\pi}{4} \right )=2\cdot \frac{\sqrt2}{2}= \sqrt{2}$

$y=r\sin\theta=2\sin\left (\frac{-\pi}{4} \right )=2\cdot \frac{-\sqrt2}{2}= -\sqrt{2}$

The rectangular coordinates are $(\sqrt2,-\sqrt2)$ .

The answer is $(\sqrt2,-\sqrt2)$ .

If , what is if is between and $2\pi$ ?

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

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

Explanation

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .