# Chapter Summary: Trigonometric Functions

• Lucretius
In summary, the problem is asking to find the values of \sin\theta and \cos\theta given the coordinates (-1,1) in a triangle with a hypotenuse of \sqrt{2}. The correct values are \sin\theta=\frac{\sqrt{2}}{2} and \cos\theta=\frac{\sqrt{2}}{2}. The mistake was made in taking the inverse of \sqrt{2} as \frac{1}{\sqrt{2}} instead of \frac{\sqrt{2}}{2}. The problem then moves on to finding the exact value of \sin\frac{5\pi}{4}, \cos90, \sin150, and \cos\frac{11
Lucretius
The problem reads: Find $\sin\theta$ and $\cos\theta$

Part a gives me the coordinates $\left(-1,1\right)$

The triangle I got had the $x-length$ as $-1$, while the $y-length$ was $1$. The hypotenuse I got was $\sqrt{2}$

Since $\sin$ is $$\frac{opposite}{hypotenuse}$$ I got $$\sin\theta=\frac{1}{\sqrt{2}}$$

The book says it is $$\sin\theta=\frac{\sqrt{2}}{2}$$

What did I mess up on? Data I'm waiting for you

Heh,he's not here.

$$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$

Do u see why?

Daniel.

dextercioby said:
Heh,he's not here.

$$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$

Do u see why?

Daniel.

Wow, I was freaking stupid. Thanks for pointing that out lol.

I think I will go hide in shame now.

Okay, Houston we have a real problem now.

It reads: Give the exact value of each expression in simplest radical form.

a. $$\sin\frac{5\pi}{4}$$ b. $\cos90$ c. $\sin150$ d. $$\cos\frac{11\pi}{6}$$

The only one I could figure out was b.

How exactly do I go about finding the radical form of these? (especially the ones in increments of $\pi$)

HINT:Use addition & subtraction formulas for sine & cosine

Daniel.

Hi!

Let's look at a. We want

$$\sin \frac{5\pi}{4} = \sin \left( \pi + \frac{\pi}{4} \right).$$

so the angle we're looking for is $\pi / 4$ past the negative x-axis (ie. it's in quadrant 3 [using the same terminology as last time], and $45^\circ$ from each axis). Does that help?

If you're able to answer that question now, try to do similar manipulations on the others to figure out where the angles are.

And dextercioby's suggestion will work just as well, if you feel like taking a more algebraic approach. Some identities that might help (and you should try to figure out for yourself why they work) follow:

$$\sin (-\theta) = -\sin \theta$$
$$\cos (-\theta) = \cos \theta$$
$$\tan (-\theta) = -\tan \theta$$
$$\sin (\pi - \theta) = \sin \theta$$
$$\cos (\pi - \theta) = -\cos \theta$$
$$\tan (\pi + \theta) = \tan \theta$$
$$\sin (\pi/2 - \theta) = \cos \theta$$
$$\cos (\pi / 2 - \theta) = \sin \theta$$

Last edited:

## What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant.

## What is the unit circle and how is it related to trigonometric functions?

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It is related to trigonometric functions because the coordinates of points on the unit circle can be used to calculate the values of trigonometric functions.

## What is the Pythagorean identity and how is it used in trigonometry?

The Pythagorean identity is sin²θ + cos²θ = 1. It is used in trigonometry to relate the values of sine and cosine to the side lengths of a right triangle.

## What are the reciprocal trigonometric functions and how are they related to the basic trigonometric functions?

The reciprocal trigonometric functions are cotangent, secant, and cosecant. They are related to the basic trigonometric functions because they are the inverses of tangent, cosine, and sine, respectively.

## How are trigonometric functions used in real-world applications?

Trigonometric functions are used in a variety of real-world applications, such as in engineering, physics, and navigation. They can be used to calculate the angles and distances of objects, as well as to model periodic phenomena and waves.

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