Solving Sin(2α)=-8/17 Using Identities

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Q2. however that is no where near the book answer of 4√17/17. the other way i get the same thing. with the same results.
  • #1
LT72884
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Homework Statement


Sin(2α)=-8/17
Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha.



Homework Equations





The Attempt at a Solution



I tried and tried BUT i am lost because in order to find cos of alpha using identies, you still have to have cos2a or cos^2a. I have not been able to find an identy that just uses sin(2a) and spits out some sort of cos. I tried using cos2a=1-2sin^2alpha but did not end up with anything close to what the book says is the answer. Funny thing is, what ever algerbra i try on the different identities, i still get the same answer that i got previously. the book says: 4√17/17, √17/17, -4, √17/4, -√17, and -1/4 but doesn't tell me if that is cos or what function it is. haha. i don't have the solutions manual either because that was an extra 50$ i couldn't afford.

any help would be nice. pretty pics IF possible but i understand if not. thanks
 
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  • #2
It's not very clear.

What are you trying to find?

If sin(θ) = -8/17, what is cos(θ) ?
 
  • #3
SammyS said:
It's not very clear.

What are you trying to find?

If sin(θ) = -8/17, what is cos(θ) ?


i wrote it exactly as the book stated it.Haha. i agree its unclear. it gave me sin(2a) and wants me to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a)

i have to use either a double angle identity or a half angle identy

i can't figure out which one to use because there is not an identy that calls for just sin. they all ask for at least cos and or sin.

thanks
 
  • #4
SammyS said:
It's not very clear.

What are you trying to find?

If sin(θ) = -8/17, what is cos(θ) ?

I think he's trying to derive the value of sin(θ), cos(θ), tan(θ), and cot(θ) using sin(2θ) = -8/17.

LT72884 said:

Homework Statement


Sin(2α)=-8/17
Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha.

Homework Equations


The Attempt at a Solution



I tried and tried BUT i am lost because in order to find cos of alpha using identies, you still have to have cos2a or cos^2a.

Yes, this is true, if you have cos(2α), then finding sin(α), cos(α), tan(α), and cot(α) should be easy as a piece of cake, right?

So the problem remains is, from sin(2α), how can you find cos(2α)? Hint: You should consider using Pythagorean Identity:

[tex]\sin ^ 2 (\gamma) + \cos^2(\gamma) = 1[/tex]

Can you go from here? :)
 
  • #5
VietDao29 said:
I think he's trying to derive the value of sin(θ), cos(θ), tan(θ), and cot(θ) using sin(2θ) = -8/17.
Yes, this is true, if you have cos(2α), then finding sin(α), cos(α), tan(α), and cot(α) should be easy as a piece of cake, right?

So the problem remains is, from sin(2α), how can you find cos(2α)? Hint: You should consider using Pythagorean Identity:

[tex]\sin ^ 2 (\gamma) + \cos^2(\gamma) = 1[/tex]

Can you go from here? :)

thanks

yeah cos(2a) is a piece of cake. a lot of writting BUT i know where to go with it. however, i spent 2 hours on this opne last night and got no where. I kept getting a different answer BUT the same answer everytime. so my results were consistant.

i have looked at that one for sin^2a=1-cos^2a but i don't end up with the same answer as they do. BUT my algerbra indeed is not that great. i still struggle with fractions. haha. if you can or would like to. do you get any of the previous answers i posted in my original post when using the Pythagorean therom?

thanks
 
  • #6
LT72884 said:
thanks

yeah cos(2a) is a piece of cake. a lot of writting BUT i know where to go with it. however, i spent 2 hours on this opne last night and got no where. I kept getting a different answer BUT the same answer everytime. so my results were consistant.

i have looked at that one for sin^2a=1-cos^2a but i don't end up with the same answer as they do. BUT my algerbra indeed is not that great. i still struggle with fractions. haha. if you can or would like to. do you get any of the previous answers i posted in my original post when using the Pythagorean therom?

thanks

Yes, I do.

Can you show me your work, so that I can check it for you?
 
  • #7
VietDao29 said:
Yes, I do.

Can you show me your work, so that I can check it for you?


dang... ok let me try it again and ill show work. thanks. give me a few please
 
  • #8
LT72884 said:
i wrote it exactly as the book stated it.Haha. i agree its unclear. it gave me sin(2a) and wants me to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a)

i have to use either a double angle identity or a half angle identity

i can't figure out which one to use because there is not an identity that calls for just sin. they all ask for at least cos and or sin.

thanks
Well, writing that you need to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a), makes it much clearer than saying "Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha."

(I see that you did end that with "alpha", at the end of that run-on string.)

If cos(2α) is a piece of cake, (except for the sign), then use cos(2x) = 2cos2(x) - 1 = 1 - 2sin2(x), to find cos(α) and sin(α)
 
  • #9
SammyS said:
Well, writing that you need to find sin(a) cos(a) csc(a) sec(a) tan(a) cot(a), makes it much clearer than saying "Using either double angle or half angle identities, sin,cos,tan,csc,sec and cot of alpha."

(I see that you did end that with "alpha", at the end of that run-on string.)

If cos(2α) is a piece of cake, (except for the sign), then use cos(2x) = 2cos2(x) - 1 = 1 - 2sin2(x), to find cos(α) and sin(α)

i agree that it was sloppy. sorry.

i have tried both of those ways and i get 13/17 or 255/289.

if sin(2a)= -8/17 then sin(a) should be 4/17, if in Q2, since sin is half of 2 alpha.

cos^2a=1-(4/17)^2 so that would be 289/289-16/289 then root that which is not anything the book gets.

using the other one.

i get this:

1-2(4/17)2

i get 1-2(16/289)

then that goes to 1-32/578

which goes to 578/578-32/578

so I am not seeing how this is getting to where i need to be.
 
  • #10
LT72884 said:
i agree that it was sloppy. sorry.

i have tried both of those ways and i get 13/17 or 255/289.

if sin(2a)= -8/17 then sin(a) should be 4/17, if in Q2, since sin is half of 2 alpha.
This is not correct. You are confusing sin(2α) with 2sin(α), which are not necessarily equal.

If sin2θ + cos2θ = 1, then surely
sin2(2α) + cos2(2α) = 1.
Substitute:
(-8/17)2 + cos2(2α) = 1
... and solve for cos(2α).

Then use the identity
cos(2α) = 2cos2(α) - 1
to find cos(α). The rest should follow.
 
  • #11
I think you said you got cos(2α). If so, what did you get for it.

cos(2α) = ___?___
 
  • #12
but isn't sin(a) half of sin(2a)? because if you go sin(2a/2) that should canle the 2's out leaving only sin(a):

so sin(2a)=-8/17

then sin(2a/2)=8/17/2 which should yield sin(a)=8/34 or 4/17

how is dividing -8/17 giving me 2sin(a)? i don't see it at all.

thanks
 
  • #13
SammyS said:
I think you said you got cos(2α). If so, what did you get for it.

cos(2α) = ___?___

i did BUt it is not what the others or the book has gotten.
 
  • #14
The problem doesn't ask for cos(2α) .
 
  • #15
LT72884 said:
but isn't sin(a) half of sin(2a)?
No, it isn't!
LT72884 said:
because if you go sin(2a/2) that should canle the 2's out leaving only sin(a):
You can't divide both sides by 2 like that.

There is another identity involving sin(2α):
sin(2α) = 2sinsin(α)cos(α)
 
  • #16
SammyS said:
The problem doesn't ask for cos(2α) .

it does because in order to get to cos(a), i need to know cos(2a) i have been trying to find cos(2a) so i can use the other identity. BUT i can't even get past the whole sin(2a) thing for some reason.
 
  • #17
sin2(x) + cos2(x) = 1

Therefore, cos(x) = √(1 - sin2(x) )
 
  • #18
eumyang said:
No, it isn't!

You can't divide both sides by 2 like that.

There is another identity involving sin(2α):
sin(2α) = 2sinsin(α)cos(α)

see now I am confused. how can 1a not be half of 2a? i don't see how dividing sin(2a) in half is wrong or not the same as sin(1a). Is it because the 2 would also be under teh sin as well since everything is attached to that?


i can't use that identity because i don't know cos(a) yet. lol
 
  • #19
LT72884 said:
it does because in order to get to cos(a), i need to know cos(2a) i have been trying to find cos(2a) so i can use the other identity. BUT i can't even get past the whole sin(2a) thing for some reason.
sin(2x) ≠ 2sin(x)
Try putting in some angles and see!
sin(45°) ≈ 0.7071
sin(90°) = 1
2 * 0.7071 ≠ 1!

SammyS, VietDao29, and I have been nudging you a certain direction. Look at my previous post #10:
eumyang said:
If sin2θ + cos2θ = 1, then surely
sin2(2α) + cos2(2α) = 1.
Substitute:
(-8/17)2 + cos2(2α) = 1
... and solve for cos(2α).

Then use the identity
cos(2α) = 2cos2(α) - 1
to find cos(α). The rest should follow.
... and please try it out.
 
  • #20
SammyS said:
sin2(x) + cos2(x) = 1

Therefore, cos(x) = √(1 - sin2(x) )

i have done that one and it doesn't work. sorry for not understanding at all but it makes no sense to me. i have tried substituting it into the normal pythagerian formula nad still get the same answer as i always get

√1-(-8/17)2

goes to:

√1-64/289
√289/289-64/289
√15/17

which is not on there
 
  • #21
LT72884 said:
i have done that one and it doesn't work. sorry for not understanding at all but it makes no sense to me.
What SammyS wanted you to do is to substitute x for 2α so that
sin(x) = -8/17,
and use the Pythagorean identity to find cos(x).

LT72884 said:
i have tried substituting it into the normal pythagerian formula nad still get the same answer as i always get

√1-(-8/17)2

goes to:

√1-64/289
√289/289-64/289
√15/17

which is not on there
First of all, it's not √15/17, it's ±15/17. Second, ±15/17 is the value of cos(2α), not cos(α). You also need to determine which value of cos(2α) is the right one to use: 15/17, or -15/17. Then, you still need to use the identity
cos(2α) = 2cos2(α) - 1
to find cos(α).
 
  • #22
eumyang said:
sin(2x) ≠ 2sin(x)
Try putting in some angles and see!
sin(45°) ≈ 0.7071
sin(90°) = 1
2 * 0.7071 ≠ 1!

SammyS, VietDao29, and I have been nudging you a certain direction. Look at my previous post #10:

... and please try it out.

i did and i still get the wrong answer

(-8/17)2+cos2(2a)=1

cos2(2a)=√1-(-8/17)2

then that goes to:

√1-64/289
√289/289-64/289
√225-289
15/17
 
  • #23
LT72884 said:
i did and i still get the wrong answer

(-8/17)2+cos2(2a)=1

cos2(2a)=√1-(-8/17)2

then that goes to:

√1-64/289
√289/289-64/289
√225-289
15/17
We seem to be crossing paths. Look at my previous post. You keep saying that you're not getting the right answer. You haven't found any answers yet. You were asked to find the 6 trig functions of alpha. Finding cos(2α) is a step towards finding the answers to this problem.

eumyang said:
What SammyS wanted you to do is to substitute x for 2α so that
sin(x) = -8/17,
and use the Pythagorean identity to find cos(x).First of all, it's not √15/17, it's ±15/17. Second, ±15/17 is the value of cos(2α), not cos(α). You also need to determine which value of cos(2α) is the right one to use: 15/17, or -15/17. Then, you still need to use the identity
cos(2α) = 2cos2(α) - 1
to find cos(α).
 
  • #24
eumyang said:
What SammyS wanted you to do is to substitute x for 2α so that
sin(x) = -8/17,
and use the Pythagorean identity to find cos(x).


First of all, it's not √15/17, it's ±15/17. Second, ±15/17 is the value of cos(2α), not cos(α). You also need to determine which value of cos(2α) is the right one to use: 15/17, or -15/17. Then, you still need to use the identity
cos(2α) = 2cos2(α) - 1
to find cos(α).

eumyang said:
sin(2x) ≠ 2sin(x)
Try putting in some angles and see!
sin(45°) ≈ 0.7071
sin(90°) = 1
2 * 0.7071 ≠ 1!

SammyS, VietDao29, and I have been nudging you a certain direction. Look at my previous post #10:

... and please try it out.

i did and i still get the wrong answer

(-8/17)2+cos2(2a)=1

cos2(2a)=√1-(-8/17)2

then that goes to:

√1-64/289
√289/289-64/289
√225-289
15/17 and its in Q3(between 180 and 270) which makes it a negative

i am really sorry guys. this is my first ever trig class and i know its frustrating for you guys. Imagine how i feel! haha. i did indeed get 15/17 for cos(2a)

now when i use cos(2a)=2*cos2a-1

i end up with :

-15/17=2*cos2a-1

add the one over

17/17-15/17=2/17

2/17=2*cos2a

now divide by 2

2/17*1/2=cos2a

2/34=cos2a

now i have

√2/34
 
  • #25
LT72884 said:
now when i use cos(2a)=2*cos2a-1

i end up with :

-15/17=2*cos2a-1

add the one over

17/17-15/17=2/17

2/17=2*cos2a

now divide by 2

2/17*1/2=cos2a

2/34=cos2a
Hold on! Reduce that fraction! Both the numerator and denominator are divisible by 2, so
1/17=cos2a

Now take the square root of both sides. You'll then need to rationalize the denominator.
 
  • #26
eumyang said:
Hold on! Reduce that fraction! Both the numerator and denominator are divisible by 2, so
1/17=cos2a

Now take the square root of both sides. You'll then need to rationalize the denominator.


DOH! dang simplifing. I always forget to do that because my professor says when drawing graphs, he does not want us to simply the fractions. IE, if we start at 0 and we need 5 points to graph a general sin wave, and our period is pi, then do everything in 4ths and don't simplify. haha, so 0,1/4,2/4,3/4,4/4. haha.

ok cool, now that i got root17/17 for cos(a)

so now i have something. now i can take

cos2a=1/17

and go

sin(a)=√1-1/17

√17/17-1/17

ends up 4/√17

OMG, this is embarrassing. haha i didnt know that you could substitute sin(2a) for sin2a in the formula I thought that that formula had to stay that way. i didnt know that you could do sin+cos=1 or sin2+cos2=1 or even sin(2a)+cos(2a)=1. See, i was informed that it had to stay sin2a+cos2a=1 had to stay that way and only manipulate the forumla around. that's why i was doing all that other bogus stuff that i was doing because i had to find sin2a or sin(a) somehow. So i take it, that as long as the sin and cos matches, it will always =1? so sin3+cos3=1

or

cos(a)cos(b)-sin(a)sin(b)+cos(a)cos(b)-sin(a)sin(b)=1 ?

thanks for all the help and being very very very very patient with me. I have never gone this far in math before. i am a mechanical engineering student. My book doesn't even say that you can manipulate the forumlas that way. so I am at the mercy of others help. haha

tahnks again guys. i think i can find the rest, if not. this thread may be bumped. haha
 
  • #27
LT72884 said:
OMG, this is embarrassing. haha i didnt know that you could substitute sin(2a) for sin2a in the formula I thought that that formula had to stay that way. i didnt know that you could do sin+cos=1 or sin2+cos2=1 or even sin(2a)+cos(2a)=1. See, i was informed that it had to stay sin2a+cos2a=1 had to stay that way and only manipulate the forumla around. that's why i was doing all that other bogus stuff that i was doing because i had to find sin2a or sin(a) somehow.
You really need to read our posts carefully. We never said that
sin θ + cos θ = 1 (don't forget to write in the angle, be it θ, x, etc.)
or that
sin(2a)+cos(2a)=1

The Pythagorean identity states:
sin2θ + cos2θ = 1
Both sine and cosine must be squared, and both angles must be the same. I earlier wrote that this was also true:
sin2(2α) + cos2(2α) = 1
This is not the same as
sin(2α) + cos(2α) = 1

LT72884 said:
So i take it, that as long as the sin and cos matches, it will always =1? so sin3+cos3=1
Nope. sin3θ + cos3θ = 1 is not an identity.
 
  • #28
eumyang said:
You really need to read our posts carefully. We never said that
sin θ + cos θ = 1 (don't forget to write in the angle, be it θ, x, etc.)
or that
sin(2a)+cos(2a)=1

The Pythagorean identity states:
sin2θ + cos2θ = 1
Both sine and cosine must be squared, and both angles must be the same. I earlier wrote that this was also true:
sin2(2α) + cos2(2α) = 1
This is not the same as
sin(2α) + cos(2α) = 1


Nope. sin3θ + cos3θ = 1 is not an identity.

so what's the difference between sin2a+cos2a=1 versus with cubes? how does this one make it an idenity but the other one not? or how does
sin2(2α) + cos2(2α) = 1 make an identity?

is it because of the squares and angles being the same that make it an identity? If so, why do squares work but not cubes? that's what confuses me is it seems like its all random. IE

sin2(2α) + cos2(2α) = 1 seems random because it doesn't look like sin2a+cos2a=1 to me at all. to me, sin3θ + cos3θ = 1 more looks like sin2a+cos2a=1 than the other one. I am just trying to learn all this and make sense of it is all. thanks

it must be because of the squares like you mentioned earlier. why are squares so important?
 
  • #29
LT72884 said:
so what's the difference between sin2a+cos2a=1 versus with cubes? how does this one make it an idenity but the other one not? or how does
sin2(2α) + cos2(2α) = 1 make an identity?

is it because of the squares and angles being the same that make it an identity? If so, why do squares work but not cubes? that's what confuses me is it seems like its all random. IE

sin2(2α) + cos2(2α) = 1 seems random because it doesn't look like sin2a+cos2a=1 to me at all. to me, sin3θ + cos3θ = 1 more looks like sin2a+cos2a=1 than the other one. I am just trying to learn all this and make sense of it is all. thanks

it must be because of the squares like you mentioned earlier. why are squares so important?

It's not all random. Are you familiar with the Pythagorean theorem from geometry:
a2 + b2 = c2
... where a and b are lengths of the two legs of a right triangle and c is the length of the hypotenuse? The Pythagorean identity
sin2θ + cos2θ = 1
is based from that theorem.

It sounds like you are not familiar with the unit circle? When you learn that, you can see more clearly why the Pythagorean identity holds.
 
  • #30
eumyang said:
It's not all random. Are you familiar with the Pythagorean theorem from geometry:
a2 + b2 = c2
... where a and b are lengths of the two legs of a right triangle and c is the length of the hypotenuse? The Pythagorean identity
sin2θ + cos2θ = 1
is based from that theorem.

It sounds like you are not familiar with the unit circle? When you learn that, you can see more clearly why the Pythagorean identity holds.

no, i know the unit circle. I spent a long time trying to figure out how they got all the sins and cos. I had to draw a 45 45 triangle with a hypotenous of 1 and then go from there to find the rest using pythagerian theory and what not. i have solved right triangles many times using the Pythagorean therom. I know its not random but sure seems like it. i googled trig identities and the sin2(2a)=cos2(2a) is not listed in my book or on goolge, so that's why it seems random. how should have i known that i could do that when all the examples in the book don't even list that as a possibility. So it seemed random because its like someone just grabbed the sin and cos functions and put them in there and set it = to 1.

thats why i asked if sin^3+cos^3=1 would work because to me that's what happened earlier, random sins and cos's were put into the pythagerian theory. haha. but sin^3+cos^3=1 does indeed work. i just did sin(90)^3+cos(90)^3 and it =1

so now I am lost again. haha.oh well. ill figure it out
 
  • #31
LT72884 said:
no, i know the unit circle. I spent a long time trying to figure out how they got all the sins and cos. I had to draw a 45 45 triangle with a hypotenous of 1 and then go from there to find the rest using pythagerian theory and what not. i have solved right triangles many times using the Pythagorean therom. I know its not random but sure seems like it. i googled trig identities and the sin2(2a)=cos2(2a) is not listed in my book or on goolge, so that's why it seems random. how should have i known that i could do that when all the examples in the book don't even list that as a possibility. So it seemed random because its like someone just grabbed the sin and cos functions and put them in there and set it = to 1.

thats why i asked if sin^3+cos^3=1 would work because to me that's what happened earlier, random sins and cos's were put into the pythagerian theory. haha. but sin^3+cos^3=1 does indeed work. i just did sin(90)^3+cos(90)^3 and it =1

so now I am lost again. haha.oh well. ill figure it out

An identity is an equality that holds true for every value of the variable(s) in that equality.

Example of an identity: sin2x + cos2x = 1. This equality holds true regardless of what value x takes. Try plugging in some random values for x to see.

--------------------------

However, sin3x + cos3x = 1 is indeed, NOT an identity. The above equality may hold for some specific value of x, but it will fail in other cases.

sin3x + cos3x = 1 will hold true for x = 90o, or even when x = 0o.

But it will FAIL to hold true, when x = 45o, and many many other values.

Regards,
 
  • #32
but sin^3+cos^3=1 does indeed work. i just did sin(90)^3+cos(90)^3 and it =1

I'm beginning to think this whole thread is a troll. If thiws goes on much longer I will close or delete this thread.
 
  • #33
HallsofIvy said:
I'm beginning to think this whole thread is a troll. If thiws goes on much longer I will close or delete this thread.

Ok sorry for a late reply. i don't start work till evening time and i don't have a pc at home for the time being.

Im beginning to understand some things about identities. i tried some random values for x and saw the results. With the cubed one i THOUGHT would work, i realized last night around 4 am that it only worked because 1 cubed is 1 and 0 cubed is, well, 0. didnt realize that till later though. I am beginning to wonder that the sqaures have something to do with it and it can only be squares and not cubes. SO what is so important about squares?

Oh and what is a troll thread? is that where someone is paid to just make random non sense on a thread?

Im sorry if it seems like that but its not. I have never done trig so for me to always remember that the sin of 90 = 1 or whatever it is is new to me. i still don't have the cos and sin memorized for all unit circle yet. haha. i have no time because homework is so demanding. haha

i was under the impression that the pythagerian theory had to stay in the form of sin2a and nothing else, hence why it seemed so random to me what you guys did. BUT I am trying to see what is going on here. Please forgive. I really really appreciate the help and time you guys have shown me. I have been a member here for a year or so and have posted other questions regarding force and mass of a 747. haha

anyway, i would like to understand why the idenitity actually works. Is it because of the squares or the fact that it uses co functions? or is it both?

Thanks

Matt

thanks
 

1. How do I solve Sin(2α)=-8/17 using identities?

To solve this equation, we can use the double angle identity for sine, which states that Sin(2α) = 2SinαCosα. We can then substitute this into the original equation to get 2SinαCosα = -8/17. From there, we can use the Pythagorean identity (Sin²α + Cos²α = 1) to solve for either Sinα or Cosα, and then use the inverse trigonometric functions to find the value of α.

2. What is the Pythagorean identity?

The Pythagorean identity is a fundamental trigonometric identity that states Sin²α + Cos²α = 1. This identity is derived from the Pythagorean theorem in geometry and is used in many trigonometric calculations.

3. Can I use a calculator to solve this equation?

Yes, you can use a calculator to solve this equation. However, it is important to make sure your calculator is set to the correct mode (degrees or radians) and that you use the inverse trigonometric functions to find the value of α.

4. Are there any other identities I can use to solve this equation?

Yes, there are other identities that can be used to solve this equation, such as the half angle identities or the sum and difference identities. However, the double angle identity is the most straightforward and efficient method for solving this specific equation.

5. What are some common mistakes when solving equations using identities?

Some common mistakes when solving equations using identities include forgetting to substitute the identity into the original equation, using the wrong identity, and not simplifying the equation before solving. It is important to carefully follow the steps and check your work to avoid these errors.

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