Question

using cos(theta)=(-3/5) , pi<0<(3pi/2) , sin=.8, find cos(2theta)

Answer 1

Sin = 0.8? or does sin = -0.8?

Answer 2

I calculated .8

Answer 3

sorry its sin(theta)=0.8

Answer 4

Ah, watch what quadrant you are in. When you take a square root, it could be the positive answer, or the negative answer. In quadrant III, both sin and cos are negative.

Answer 5

So, you know that cos(theta) = -3/5, and sin(theta) = -0.8.

We want to know cos(2theta). What identity would we use?

Answer 6

adding??

Answer 7

lol i'm not sure.. my teacher isn't the best.

Answer 8

\[\cos(2\Theta) = \cos^2(\Theta)-\sin^2(\Theta) = 2\cos^2(\Theta) - 1\]

Answer 9

so, for 2cos^2(theta)-1, i'd plug in (-3/5)?

Answer 10

Yeah. We'll check the sign of your result after you do that.

Answer 11

ohkay one second

Answer 12

how do i get cos^2 on my calculator

Answer 13

You know cos(theta) = -3/5

So cos^2(theta) = (-3/5)^2 = 9/25

You can just put this value in for cos^2(theta)

Answer 14

-.28 ?

Answer 15

That's what I get :)

Answer 16

ok thankyou so much. one more question?

Answer 17

What quadrant do you expect to "land" in? Is the sign correct on your value for cos(2*theta)?

Answer 18

I would expect to land in 2nd?

Answer 19

because it says find sin(theta) which i did and found .8

Answer 20

so , it'd be like (-2.8, .8)

Answer 21

sin(theta) is -0.8, as we start in the third quadrant, where both cosine and sine are negative.

Answer 22

why do we start in third quadrant?

Answer 23

but yes, we expect to land in quadrant II, where cos is negative.

We start in quadrant III because we are between pi and 3pi/2, which is quadrant III

Answer 24

Oh because pi<0<-3pi/2 Ohkay ! yay ! haha thankyou.

Answer 25

well can you help me find sin(2pi/3-theta)

Answer 26

Since the absolute value of sin > absolute value of cos, we are closer to 3pi/2 than to pi. When we double the angle, we will land in quadrant II, cos is negative, and all is well.

Answer 27

Well, the first thing we can do is to put that into an easier form. We know that sin(-x) = -sin(x).

So,

\[\sin(\frac{2 \pi}{3} - \Theta) = \sin(-(\Theta - \frac{2\pi}{3})) = -\sin(\Theta - \frac{2\pi}{3})\]

Let me think of how to get a numerical value.

Answer 28

So, you can do it with a calculator, if you are careful. But I can't come up with a good way to do it without one.

Answer 29

its ohkay, i'm not really sure.. ohkay , so for -sin(theta-2pi/3) i can plug in -.8 for theta, and get the answer?

Answer 30

No, -0.8 is the value for sin(theta). You'd need to find the actual angle.

This is complicated slightly because calculators only report values for sin and cos in two quadrants. cos is reported in quadrants 1 and 2. sin in quadrants 1 and 4.

Answer 31

ohhh im so confused ! haha ohkay thankyou.. could you help me with something else

Answer 32

I can sure try :)

Answer 33

this should be easier.. it says find all possible answers for sin(x)+cos(2x)=1

Answer 34

Much easier :)

We can use either the double angle formula or half angle formula to get all the trig functions to be of the same type.

cos(2x) = 1-2sin^2(x)

Therefore, sinx+1-2sin^2(x) = 1

Sin(x) - 2sin^2(x) = 0

Sin(x) = 2Sin^2(x)

Sin(x) = 1/2

Answer 35

ohkay, and i think this is multiplied sin*cos(2x)+cos*sin(2x)=0

Answer 36

On the last one, make sure you find the values for x

Answer 37

it doesn't give me values for x

Answer 38

it just says to find possible answers..

Answer 39

Right. The possible answers are all the values of x that make sin(x) equal to 1/2

Answer 40

ohhhh , how do i get that

Answer 41

Well, you can find one with a calculator easily. The other one is a related angle in another quadrant (sine is positive in quadrants I and II)

Answer 42

For the last question you asked, we do something similar.
Using two identities:
\[Cos(2x) = 2Cos^2(x) - 1\]
and
\[Sin(2x) = 2Sin(x)Cos(x)\]

We can rewrite the equation as:
\[Sin(x)*(2Cos^2(x) - 1) + Cos(x)(2Cos(x)Sin(x)) = 0\]
Which simplifies to:
\[2Sin(x)Cos^2(x) - Sin(x) + 2Sin(x)Cos^2(x) = 0\]
We add sin(x) to each side, and then divide by sin(x), as well as adding the other two terms together, and arrive at:
\[4Cos^2(x) = 1\]
\[Cos^2(x) = \frac{1}{4}\]
\[Cos(x) = \frac{1}{2}\]

Find the values for x that make this true.