sin (theta/2) cos (theta/2) tan (theta/2)

[nathanralph]

So I'm trying to solve this problem and for some reason the online program keeps telling me this is incorrect.

I am to find sin (theta/2) cos (theta/2) tan (theta/2).

sec (theta) = -4 180degrees<theta<270degrees

I drew a triangle picture and got that:

hyp= -4
opp= sqr(15)
adj= 1

So I attempt to solve the problem.

I come up with sqr(10)/4 for sin(theta/2) fine, and it tells me that is correct.

But I run into troubles finding cos, and tan.

here are my steps for sin:

sqr[1-(-1/4)/2] = sqr[(1+1/4)/2] = sqr[(5/4)/2] = sqr(5/8) = [sqr(40)]/8 = [2sqr(10)]/8 = [sqr(10)]/4

Obviously then, tan is sin/cos, but I'm going to get tan wrong if I get cos wrong. Is there something wrong with my calculations? I've done this about 6 times with the same result.

 

[tkhunny]

Why do you need to find cosine and tangent?

sin(x)*cos(x)*tan(x) = sin(x)*cos(x)*[sin(x)/cos(x)] = [sin(x)]²

I didn't check the rest of your arithmetic.

 

[nathanralph]

Well, the problem instructs me to find cos and tan.

So, are you saying that if I take sin^2 I can work backwards and get cos and tan?

 

[Denis]

cos(whatever) = sqrt(1 - sin(whatever)^2)

tan(whatever) = sin(whatever) / cos(whatever)

etch that in your memory cells

 

[nathanralph]

cos(whatever) = sqrt(1 - sin(whatever)^2)

tan(whatever) = sin(whatever) / cos(whatever)

etch that in your memory cells

yeah...I understand those, but in this case I'm using half-angle formulas.

 

[Guest]

but the angle can be whatever as long as it is the same unknown.
Replace your theta/2 with alpha and look at the problem again.

 

[tkhunny]

Well, the problem instructs me to find cos and tan.

So, are you saying that if I take sin^2 I can work backwards and get cos and tan?

No, I'm saying, once you have sine, you are done. cosine and tangent are unnecessary.

If your problem statement has requirements, PLEASE tell us what they are.

Volunteers, raise your hand if you can read minds...{waiting}...{waiting}...Nope. No volunteers.

See, you'll just have to tell us how we get to proceed if the problem statement provides specifics.

 

[nathanralph]

Ok I solved it that way. Problem is - I get the exact same answer that the computer tells me is incorrect.

cos X = sqr(1-sin^2 X)
cos X = sqr(1-(sqr(10)/4)^2)
cos X = sqr[1-(10/16)]
cos X = sqr(3/8)
cos X = sqr(3*8)/8
cos X = sqr(24)/8
cos X = 2sqr(6)/8
cos X = sqr(6)/4

 

[nathanralph]

Well, the problem instructs me to find cos and tan.

So, are you saying that if I take sin^2 I can work backwards and get cos and tan?

No, I'm saying, once you have sine, you are done. cosine and tangent are unnecessary.

If your problem statement has requirements, PLEASE tell us what they are.

Volunteers, raise your hand if you can read minds...{waiting}...{waiting}...Nope. No volunteers.

See, you'll just have to tell us how we get to proceed if the problem statement provides specifics.

I did.

"I am to find sin (theta/2) cos (theta/2) tan (theta/2)."

Sorry if that was too vague....perhaps I should state it differently.

 

[stapel]

"I am to find sin (theta/2) cos (theta/2) tan (theta/2)."

And the tutors have explained how to find this product:

sin(x)*cos(x)*tan(x) = sin(x)*cos(x)*[sin(x)/cos(x)] = [sin(x)]²

Plug your value for "sin(x)", and use provided solution.

Eliz.

 

[nathanralph]

And the tutors have explained how to find this product:

Yeah...I know. And I do appreciate it. I was just trying to clarify it.

I think there is something wrong with the computer program. I have used all of the methods here, and come up with the correct answer...I think there's a flaw in this particular problem online.

In any case, Thanks to all who helped me.