Find theta given known cos and sin....

[Helly123]

Homework Statement

Homework Equations

cos 2theta = costheta^2 - sintheta^2

The Attempt at a Solution

cos2theta = 1
2theta = 0, 2phi

but i get wrong answer.. how is it?

 

[Ray Vickson]

Homework Statement

View attachment 205960

Homework Equations

cos 2theta = costheta^2 - sintheta^2

The Attempt at a Solution

cos2theta = 1
2theta = 0, 2phi

but i get wrong answer.. how is it?

Well, given your values for $\sin \theta$ and $\cos \theta$ you should NOT get $\cos 2 \theta = 1$. Check your algebra.

 

[Helly123]

Well, given your values for $\sin \theta$ and $\cos \theta$ you should NOT get $\cos 2 \theta = 1$. Check your algebra.

oh.. cos 2 theta = root 2 / 2
2theta = 45degrees
the sin is negative, theta must on quadrant 2 or 3
225 or 315 degrees.

 

[phinds]

oops. never mind

 

[mjc123]

π

You get the "wrong" answer because the problem posits an impossible condition. Just use arccos and arcsin and you'll see they are talking about two different angles so it can't be right to call them both the same. For that matter, you don't even have to do any math; the signs alone tell you the angles are different.

No, with cos +ve and sin -ve, the angle must be in the 4th quadrant. All silly tom cats. (A calculator will give you different angles.)
θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 (note the question asked for the answer as a multiple of π, not in degrees).

 

[phinds]

No, with cos +ve and sin -ve, the angle must be in the 4th quadrant. All silly tom cats. (A calculator will give you different angles.)
θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 (note the question asked for the answer as a multiple of π, not in degrees).

Yes, I had already deleted my post before you posted this

 

[Ray Vickson]

π
No, with cos +ve and sin -ve, the angle must be in the 4th quadrant. All silly tom cats. (A calculator will give you different angles.)
θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 (note the question asked for the answer as a multiple of π, not in degrees).

Or, the angle could be $-\pi/4$, assuming negative angles are allowed.

 

[mjc123]

The question said 0 ≤ θ < 2π.

 

[Helly123]

But the answer also 15/4 pi.. Which is 337.5 degree

 

[Ray Vickson]

But the answer also 15/4 pi.. Which is 337.5 degree

No, it isn't. $2 \pi \leftrightarrow 360^o$, and $15/4> 3> 2$, so $15/4 \pi$ is a lot bigger than $360^0$.

 

[Helly123]

no

No, it isn't. $2 \pi \leftrightarrow 360^o$, and $15/4> 3> 2$, so $15/4 \pi$ is a lot bigger than $360^0$.

the key answer it is. the 2 theta allowed to be bigger than 2 pi. just theta < 2pi

 

[SammyS]

no

the key answer it is. the 2 theta allowed to be bigger than 2 pi. just theta < 2pi

Yes, but you said:

But the answer also 15/4 pi.. Which is 337.5 degree

As Ray said, that's not true. After all, (15/4)π ≠ 337.5° .

Perhaps you meant that

one of the answers is 2θ = (15/4)π , which means that θ = 337.5° ,​

so, of course, θ = (15/8)π < 2π .

 

[MidgetDwarf]

But the restrictions say theta is defined on [0,2pie)?
Hence, 2theta=-pie/4.

From our restriction, this -pie/4 should be equal to 7pie/4 (sin is negative and cos is positive/ IV Quadrant).

Therefore 2theta=7pie/4
theta=7pie/8

 

[Mark44]

But the restrictions say theta is defined on [0,2pie)?
Hence, 2theta=-pie/4.

From our restriction, this -pie/4 should be equal to 7pie/4 (sin is negative and cos is positive/ IV Quadrant).

Therefore 2theta=7pie/4
theta=7pie/8

You do know that there is a difference between "pie" and "pi" with the latter being the name of a Greek letter, right?

 

[MidgetDwarf]

You do know that there is a difference between "pie" and "pi" with the latter being the name of a Greek letter, right?

typing with a new tablet... with auto correct...

You do know that making a post within a topic, that is not relevant to the discussion is considered trolling? That is a violation of PF rules of conduct.

 

[mjc123]

But the restrictions say theta is defined on [0,2pie)?
Hence, 2theta=-pie/4.

From our restriction, this -pie/4 should be equal to 7pie/4 (sin is negative and cos is positive/ IV Quadrant).

Therefore 2theta=7pie/4
theta=7pie/8

The restriction applies to θ, not 2θ. It is also θ that must be in the fourth quadrant, therefore θ = 7π/8 is not acceptable

 

[ehild]

The restriction applies to θ, not 2θ. It is also θ that must be in the fourth quadrant, therefore θ = 7π/8 is not acceptable

The restriction applies to theta, but cos(2θ) is positive and sin(2θ) is negative, therefore 2θ is in the fourth quadrant or 2pi more. And the problem asks 2θ.

 

[mjc123]

2θ is indeed in the fourth quadrant, but so is θ, as cos θ is positive and sin θ negative, so 2θ must be in the "8th quadrant", as it were.

 

[Helly123]

@mjc123 why θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 ?
Why you substract π/8 from 2π?

 

[Helly123]

π
No, with cos +ve and sin -ve, the angle must be in the 4th quadrant. All silly tom cats. (A calculator will give you different angles.)
θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 (note the question asked for the answer as a multiple of π, not in degrees).

@mjc123 2 θ = is 45 degrees, theta = 22.5 degrees
So quadrant 4 is 360-22.5 = 337.5
So 2theta = 337.5 x 2 = 675 degrees (15/4)pi
Why can't i use directly 45degrees, so theta in quadrant 4 is 360-45 = 315 degrees
Since 315 and 675 have same cos value, why the answer is 675?

 

[Helly123]

Never mind i get it. Thanks for all help

 

[mjc123]

@mjc123 why θ = 2π - π/8 = 15π/8, and 2θ = 15π/4 ?
Why you substract π/8 from 2π?

From the magnitudes of cosθ and sinθ, the basic angle is 22.5° (π/8). (If you have to do this without a calculator, from the magnitude of cos 2θ, 2θ = 45°.) However, since cosθ is positive and sinθ is negative, θ must be in the 4th quadrant, i.e. between 3π/2 and 2π.

Why can't i use directly 45degrees, so theta in quadrant 4 is 360-45 = 315 degrees
Since 315 and 675 have same cos value, why the answer is 675?

I assume you mean "2 theta in quadrant 4 is 360-45". You can't use this because (for the millionth time) θ must be in the 4th quadrant. So 2θ must be between 3π and 4π.

 

[Helly123]

From the magnitudes of cosθ and sinθ, the basic angle is 22.5° (π/8). (If you have to do this without a calculator, from the magnitude of cos 2θ, 2θ = 45°.) However, since cosθ is positive and sinθ is negative, θ must be in the 4th quadrant, i.e. between 3π/2 and 2π.

I assume you mean "2 theta in quadrant 4 is 360-45". You can't use this because (for the millionth time) θ must be in the 4th quadrant. So 2θ must be between 3π and 4π.

Haha yes. Thanks

 

[Mark44]

You do know that making a post within a topic, that is not relevant to the discussion is considered trolling?

At PF we endeavor to maintain a high-quality site for the pursuit of science and mathematics, one aspect of which is to not conflate the number $\pi$ (or pi) with the dessert whose name is pronounced the same. I am aware that you have a strong knowledge of mathematics, but because you wrote "pie" six times in your post, I believed that this deserved comment and was relevant to the discussion.

That is a violation of PF rules of conduct.

As a mentor, I am very familiar with the forum rules. Hijacking a thread is a violation, but comments that relate directly to posts in the thread are neither off-topic nor hijacks.

 

[MidgetDwarf]

The restriction applies to θ, not 2θ. It is also θ that must be in the fourth quadrant, therefore θ = 7π/8 is not acceptable

Thank you. I now fully understand. So silly to make that common mistake.

 

[Helly123]

If i used: sin2x = 2sinx cos x
Sin 2x = $\frac{ -\sqrt2}{2}$
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?

 

[SammyS]

If i used: sin2x = 2sinx cos x
Sin 2x = $\frac{ -\sqrt2}{2}$
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?

You can reduce those fractions. 45/36 = 5/4, etc., and it's pi, not phi.

More to the point for your question here:

In the problem statement the requirement is, 0 ≤ x < 2π . (Actually it has θ, not x. That's not important.)

Therefore, 0 ≤ 2x < 4π. Right?

So find the additional solutions which fit that. (Add 2π to those two answers you have for 2x .)

By The Way:
In the blue strip above the box in which you compose posts, there is a large ∑ character. Click on that to find a large set of characters useful for math/science.
θ π φ ∅ ℝ etc.

 

[Helly123]

X (theta) must be on 4 Quadrant, so 2x must on 8 quadrants, 540<2x<=720
X 1=5/4 π
X2 = 7/4 π
Both added 2π to get x at 8 quadrants
X1 = 13/4 π
X2 = 15/4 π

Both satisfied the domain
How to decide which one is the answer?

 

[Helly123]

I'm confused, why cos 13/4 π is negative, even though 13/4 π on 8 quadrants?

 

[Ray Vickson]

If i used: sin2x = 2sinx cos x
Sin 2x = $\frac{ -\sqrt2}{2}$
So 2x = 45/36 phi, 63/36 phi?
X = 45/72 phi or x = 63/72 phi
While x must on 4th quadrant..
Both x above not on 4th Q ..
What's wrong ?

Because of "circularity", we have:
1st quadrant ↔ 5th quandrant ↔ 9th quandrant, etc.
4th quadrant ↔ 8th quadrant ↔ 12th quadrant, etc.
(Here, ↔ means "geometrically equivalent".)

Get it? The are only 4 physically different quadrants, so the "8th quadrant" occupies the same physical location as the 4th quadrant, and so forth. That is why both $\theta$ and $2 \theta$ can be in the same (4th = 8th = 12th...) actual, geometric quadrant.

 

[SammyS]

X (theta) must be on 4 Quadrant, so 2x must on 8 quadrants, 540<2x<=720
X 1=5/4 π
X2 = 7/4 π
Both added 2π to get x at 8 quadrants
X1 = 13/4 π
X2 = 15/4 π

Both satisfied the domain
How to decide which one is the answer?

Those actually should written be something like

2x₁ = (5/4) π
2x₂ = (7/4) π​

Then adding 2π gives (two new possibilities for x):

2x₃ = (13/4) π
2x₄ = (15/4) π​

Solve for each x_k .