Question

find the exact values of sin 2u and cos 2u using double angle formulas.
csc u = 3, pi/2< u < pi

Answer 1

\(\csc(u) = 3 ==> \sin(u) = 1/3\) in Quadrant II

What now?

Answer 2

find the exact value of sin2u and cos2u using double angle formula
cos u = -2/3, pi/2<u<pi

Answer 3

You have to know the basic relationships, like \(\sin(x) = 1/\csc(x)\). If you do not know these, you must learn them before you can solve such problems.

You must also know where things are positive and negative. The sine is positive in Quadrants I and II. The cosine is positive in Quadrants IV and I. Things like that.

Answer 4

I actually just messaged you talking about my understanding, if you could reply to my message that would be nice :) Where would I get the basics from? @tkhunny

Answer 5

Your on-line course offers NOTHING?

Answer 6

Nothing at all, Ive tried youtube, khan academy and it doesn't show me the steps, so basically Im just taking quizzes and not knowing how to solve them @tkhunny

Answer 7

Awesome.

There are no steps to memorizing basic information. You just have to memorize it.

First, most important question... Why are you registered for this course? How did you get in? Was there a placement exam? How did we get where we are - with you thrown into the shark tank?

Answer 8

I have to have the course to graduate. No there was no placement test, but because I was in a classroom setting learning then I got moved to a online school its totally different. Things I see online Ive never seen in the classroom so its all new to me.

Answer 9

Super.

Here's a place to start: http://www.regentsprep.org/Regents/math/algtrig/ATT5/unitcircle.htm

This is some of the basic information that needs to be memorized. Can we assume you are familiar with an x-y coordinate plane?

Answer 10

talking about graphs right?

Answer 11

Right. The upper-right corner is where y > 0 and x > 0. It is called Quadrant I. For trig, we chop things up a little differently. We have sin(u) > 0 and cos(u) > 0 where u is an angle. The angle 0 or 0º is the positive x-axis. The angle pi/2 or 90º is the positive y-axis. Everything in-between those two is just another example of Quadrant I.

Making any sense at all, yet?

Answer 12

Im kinda understanding a little better. Im actually looking at the website you gave me and taking notes.

Answer 13

That giant circle just a little down from the top is of utmost importance. You will need 0 (0º), pi/6 (30º), pi/4 (45º), pi/3 (60º), and pi/2 (90º) until ALL the cows come home.

Answer 14

So to solve my questions I have I need to write those down?

Answer 15

then like the other questions are like given 90 degrees, < x < 180 degrees and cos x = -3/5, find the value of sin 2x @tkhunny

Answer 16

cos(x) = -3/5 refers to an angle, x, that lies in either Quadrant II or Quadrant III. It cannot be in Quadrants I or IV. You need to see that, first.

Answer 17

Okay Ill try to understand that.

Answer 18

You also need the Pythagorean Theorem. Do you have that? We'll use that A LOT!

Answer 19

I need that.

Answer 20

Yup. That's the very first section in the link I gave you. It discusses the relationship of the Unit Circle, Right Triangles, the trigonometric functions, and the Pythagorean Theorem.

Answer 21

Okay. Ive looked into it.

Answer 22

Next, we need the concept of a "Reference Angle". That Unit Circle relates DIRECTLY to Right Triangles in Quadrant I. It's just a little bit trickier an the other three quadrants.

Answer 23

Could you explain?

Answer 24

Do you see that drawing at the top of the link? It helps establish waht we call the Pythagorean Identity.

Pythagorean Theorem (Right Triangles)
\(a^{2} + b^{2} = c^{2}\)
Pythagorean Theorem (Right Triangle on x-y Coordiante Axes)
\(x^{2} + y^{2} = r^{2}\) - Where 'r' is the distance of the point from the Origin.


Pythagorean Identity (Unit Circle)
\(cos^{2}(\alpha) + \sin^{2}(\alpha) = 1\)

You must stick all this in your head. Fundamental Importance.

Answer 25

its written down w my notes :)

Answer 26

Will these help me solve those type of problems i needed help with.

Answer 27

1. Find the exact value of sin 2u and cos 2u using the double angle formula
csc u = 3, pi/2 < u < pi

2. Rewrite 4-8 sin^2 x using a double angle formula

3. Find the exact value of sin 2u and cos 2u using the double angle formulas
cos u = -2/3, pi/2 < u < pi

4. Given 0 degrees < x < 90 degrees and sin x = 1/4, find sin 2x

5. Given 270 degrees < x < 360 degrees and cos x = 2/5, find the exact value of sin 2x

6. Given 0 degrees < x < 90 degrees and sin x = 1/4 , find cos 2x

7. Given 270 degrees< x< 360 degrees and cos x =2/5, find the exact value of cos 2x

8. Given sin x = 3/4 and lies in Quadrant 1 , find sin 2x

9. Given 90 degrees < x < 180 degrees and cos x = -3/5, find the exact value of sin 2x

Answer 28

You'll have to find the Double Angle Formulas.

Answer 29

How do you do that? I have to turn it in by 12 tonight and I need help.

Answer 30

Formulas expressing trigonometric functions of an angle 2x in terms of functions of an angle x,

sin(2x) = 2sinxcosx
(1)
cos(2x) = cos^2x-sin^2x
(2)
= 2cos^2x-1
(3)
= 1-2sin^2x
(4)
tan(2x) = (2tanx)/(1-tan^2x).
(5)
The corresponding hyperbolic function double-angle formulas are

sinh(2x) = 2sinhxcoshx
(6)
cosh(2x) = 2cosh^2x-1
(7)
tanh(2x) = (2tanhx)/(1+tanh^2x).


this is what I found on the double angle formulas.