# Can a direction cosine squared be negative?

#### Krushnaraj Pandya

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1. The problem statement, all variables and given/known data
A line makes some angle T with each of x and z axis, and angle U with y axis so that sin^2(U)=3sin^2(T).
2. Relevant equations
2cos^2(T)+cos^2(U)=1 .......(i)
cos^2(x)+sin^2(x)=1

3. The attempt at a solution
Using the above two eqns. give us the correct value for cos^2(T) which is 3/5 as per my textbook but plugging it back into (i) gives us cos^2(U)= -1/5. Am I wrong somewhere or is it a mistake in the book?

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#### PeroK

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1. The problem statement, all variables and given/known data
A line makes some angle T with each of x and z axis, and angle U with y axis so that sin^2(U)=3sin^2(T).
2. Relevant equations
2cos^2(T)+cos^2(U)=1 .......(i)
cos^2(x)+sin^2(x)=1

3. The attempt at a solution
Using the above two eqns. give us the correct value for cos^2(T) which is 3/5 as per my textbook but plugging it back into (i) gives us cos^2(U)= -1/5. Am I wrong somewhere or is it a mistake in the book?
How are you getting $\cos^2 T = \frac35$? Clearly that is too large, given your first equation.

#### SammyS

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1. The problem statement, all variables and given/known data
A line makes some angle T with each of x and z axis, and angle U with y axis so that sin^2(U)=3sin^2(T).
2. Relevant equations
2cos^2(T)+cos^2(U)=1 .......(i)
cos^2(x)+sin^2(x)=1

3. The attempt at a solution
Using the above two eqns. give us the correct value for cos^2(T) which is 3/5 as per my textbook but plugging it back into (i) gives us cos^2(U)= -1/5. Am I wrong somewhere or is it a mistake in the book?
Of course for $\ \cos(U) \$ having a real number value, its square cannot be negative. Something's wrong.

#### mjc123

Science Advisor
The maximum value of sin2 U is 1, when U = 90°. Then T is ±45° and sin2 T =1/2, so your initial equation must be wrong.

#### Krushnaraj Pandya

Gold Member
How are you getting $\cos^2 T = \frac35$? Clearly that is too large, given your first equation.
sin^2(U)=3sin^2(T). is given- change it to cos^2 and it gives us this

#### Krushnaraj Pandya

Gold Member
Of course for $\ \cos(U) \$ having a real number value, its square cannot be negative. Something's wrong.
Must've been a mistake in the book, thank you very much

#### Krushnaraj Pandya

Gold Member
The maximum value of sin2 U is 1, when U = 90°. Then T is ±45° and sin2 T =1/2, so your initial equation must be wrong.
Alright, Thank you for your help :D

#### SammyS

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Must've been a mistake in the book, thank you very much
Does the book give a result for $\ \cos(U) \,,\text{or } \cos^2(U) \,?$

#### PeroK

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sin^2(U)=3sin^2(T). is given- change it to cos^2 and it gives us this
Where did the $3/5$ come from?

#### Krushnaraj Pandya

Gold Member
Where did the $3/5$ come from?
Changing the given equation in sin to cos we get 3cos^2(T)-cos^2(U)=2. Using 2Cos^2(T)+Cos^2(U)=1 we have 2 equations in two variables, adding both gives cos^2(T)=3/5

#### Krushnaraj Pandya

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Does the book give a result for $\ \cos(U) \,,\text{or } \cos^2(U) \,?$
The book says cos^2(T)=3/5. which gives a negative value of cos^2(U)

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#### Krushnaraj Pandya

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Does the book give a result for $\ \cos(U) \,,\text{or } \cos^2(U) \,?$
Where did the $3/5$ come from?
The maximum value of sin2 U is 1, when U = 90°. Then T is ±45° and sin2 T =1/2, so your initial equation must be wrong.
There seems to be another anomaly, which is perhaps connected- there is a mcq saying- "A line OP through origin is inclined at 30 and 45 degrees respectively to OX and OY. The angle at which it is inclined to OZ is given by?" And the correct option given is "Not any angle" even though using the same equation ∑cos^2(θ)=1 gives us 3/4 + 1/2 +cos^2(x)=1 which again gives a negative value of cos^2(x)

#### PeroK

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There seems to be another anomaly, which is perhaps connected- there is a mcq saying- "A line OP through origin is inclined at 30 and 45 degrees respectively to OX and OY. The angle at which it is inclined to OZ is given by?" And the correct option given is "Not any angle" even though using the same equation ∑cos^2(θ)=1 gives us 3/4 + 1/2 +cos^2(x)=1 which again gives a negative value of cos^2(x)
Okay, I understand now. You can see that the two cones implied by those angles do not intersect, so no such line exists.

#### Krushnaraj Pandya

Gold Member
Okay, I understand now. You can see that the two cones implied by those angles do not intersect, so no such line exists.
Is that the case for the original question as well? Or perhaps they just forgot this relation and only wanted to test if I knew sum of direction cosines is 1? Because the observation that one of it is negative is a secondary one derived after getting the answer in the first place and wasn't noticed by the question-setter

• PeroK

#### PeroK

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Is that the case for the original question as well? Or perhaps they just forgot this relation and only wanted to test if I knew sum of direction cosines is 1? Because the observation that one of it is negative is a secondary one derived after getting the answer in the first place and wasn't noticed by the question-setter
Your guess is as good as mine.

• SammyS

#### Krushnaraj Pandya

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Your guess is as good as mine.
Thank you very much, I am quite sure now that that is the case

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