Proving tan\theta=(1-\sqrt{2})cot\theta

  • Thread starter ron_jay
  • Start date
In summary: I'm not sure what you're trying to prove. However, if you can find an equation that relates tan(\theta) to cot(\theta), then you can solve for \theta.
  • #1
ron_jay
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Homework Statement



There are two inclined planes each having an inclination of [tex]\alpha[/tex]. A projectile is launched from the midpoint between these two inclines such that the distance from the projectile on the ground to the inclines is x on either side.The angle of projection is [tex]\alpha[/tex] + [tex]\theta[/tex]and the initial velocity is u. The projectile just grazes the lower incline(having acute angle or [tex]\alpha[/tex] towards it) and lands on the other incline with an angle of 90 degrees to the incline.

Homework Equations



how do we prove that :

tan [tex]\theta[/tex]=(1-[tex]\sqrt{2}[/tex])cot [tex]\theta[/tex]

The Attempt at a Solution



Rotate the axis by [tex]\alpha[/tex] along the incline at the midpoint so that we have the angle of projection as [tex]\theta[/tex].The component acting downwards instead of g would be gcos [tex]\theta[/tex] and backwards gsin [tex]\theta[/tex].Next we equate it.Right?
 
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  • #2
ron_jay said:

Homework Statement



There are two inclined planes each having an inclination of [tex]\alpha[/tex]. A projectile is launched from the midpoint between these two inclines such that the distance from the projectile on the ground to the inclines is x on either side.The angle of projection is [tex]\alpha[/tex] + [tex]\theta[/tex]and the initial velocity is u. The projectile just grazes the lower incline(having acute angle or [tex]\alpha[/tex] towards it) and lands on the other incline with an angle of 90 degrees to the incline.

Homework Equations



how do we prove that :

tan [tex]\theta[/tex]=(1-[tex]\sqrt{2}[/tex])cot [tex]\theta[/tex]

The Attempt at a Solution



Rotate the axis by [tex]\alpha[/tex] along the incline at the midpoint so that we have the angle of projection as [tex]\theta[/tex].The component acting downwards instead of g would be gcos [tex]\theta[/tex] and backwards gsin [tex]\theta[/tex].Next we equate it.Right?

Good idea to rotate the axes... I'm guessing there's supposed to be an [tex]\alpha[/tex] in that equation you need to prove. also the downward component of gravity would be [tex]gcos\alpha[/tex] and the backward would be [tex]gsin\alpha[/tex]

Not sure what you meant by equate it... but treat it like an ordinary kinematics problem now... see what you can get from the kinematics equations. Note that when the object lands, the backward component of velocity is 0 (along the new axes)...
 
  • #3
i didn't get the question.probablya diagram should help.
well if the projectile is launcehd in between them how cum it touches both of them?
 
  • #4
Good idea to rotate the axes... I'm guessing there's supposed to be an [tex]\alpha[/tex].

Not sure what you meant by equate it... but treat it like an ordinary kinematics problem now... see what you can get from the kinematics equations. Note that when the object lands, the backward component of velocity is 0 (along the new axes)...

i didn't get the question.probablya diagram should help.
well if the projectile is launcehd in between them how cum it touches both of them?

Here's a diagram which should clear your understanding.
 

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  • #5
ron_jay said:
Here's a diagram which should clear your understanding.

Are you sure that's what you're supposed to prove? I went through the problem and I get that relationship but I get [tex]cot\alpha[/tex] on the right side.
 

1. What is the equation "Proving tanθ=(1-√2)cotθ" trying to prove?

The equation is trying to prove that the tangent of an angle (θ) is equal to the product of (1-√2) and the cotangent of the same angle.

2. How do you prove this equation mathematically?

One way to prove this equation is by starting with the definition of tangent and cotangent in terms of sine and cosine, and then using algebraic manipulations to simplify the expression until it is in the form of (1-√2)cotθ.

3. Can you provide an example to help understand this equation better?

Sure, let's say that θ is equal to 45 degrees. Using a calculator, we can find that the tangent of 45 degrees is 1 and the cotangent is also 1. Plugging these values into the equation, we get 1=(1-√2)*1, which is a true statement.

4. Why is this equation important in mathematics?

This equation is important because it shows a relationship between two trigonometric functions, tangent and cotangent, and it can be used to solve various problems in geometry, physics, and engineering.

5. Are there any real-life applications of this equation?

Yes, this equation can be used in various real-life situations, such as calculating the height of an object using trigonometric functions, finding the slope of a road or hill, and determining the angle of inclination for building structures.

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