cos2(x) + tan(x)cos2(x) = k

[Some1]

I have to work out x:

cos²(x) + tan(x)cos²(x) = k

in terms of k where k is a constant.

Thanks in advance.

 

[Ishuda]

I have to work out x:

cos²(x) + tan(x)cos²(x) = k

in terms of k where k is a constant.

Thanks in advance.

What work have you done so far? Where are you stuck?

Oh, and you realize that only a restricted set of k will work for real valued x, don't you?

 

[Some1]

I know, I have actually simplified the question a bit just to make it easier to answer.
The actual equation looks something more like:

h*cos²(x) + d*tan(x)cos²(x) = g*d*0.5/v²

Where every variable is known except for x.
This equation should not always have an answer.

 

[Some1]

Trig Question

I have to work out x:

h*cos²(x) + d*tan(x)cos²(x) = g*d*0.5/v²

in terms of k,h and d where k, h and d are constants.

Thanks in advance.

 

[Deleted member 4993]

I have to work out x:

h*cos²(x) + d*tan(x)cos²(x) = g*d*0.5/v²

in terms of k,h and d where k, h and d are constants.

Thanks in advance.

Yes... go at it....

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[stapel]

I have to work out x:

h*cos²(x) + d*tan(x)cos²(x) = g*d*0.5/v²

in terms of k,h and d where k, h and d are constants.

Note that tan(x) = sin(x)/cos(x), so you have:

. . . . .h cos²(x) + d cos(x) = 0.5gd/v²

. . . . .h cos²(x) + d cos(x) - 0.5gd/v² = 0

This is a quadratic in cos(x). Solve with the Quadratic Formula. Then solve the resulting trig equation(s).

 

[Ishuda]

I know, I have actually simplified the question a bit just to make it easier to answer.
The actual equation looks something more like:

h*cos²(x) + d*tan(x)cos²(x) = g*d*0.5/v²

Where every variable is known except for x.
This equation should not always have an answer.

Hint 1: If you were to plot a cos(t) + b sin(t) [with not both a and b equal to zero], you would notice that it also looked like a sine wave, maybe shifted a bit and with a greater (or smaller) amplitude that a normal sine wave by itself. That's becase you can re-write the expression in the following way: Let
c = \(\displaystyle \sqrt{a^2 + b^2}\)
sin\(\displaystyle (\theta) = \frac{a}{c}\)
cos\(\displaystyle (\theta) = \frac{b}{c}\)
Then
a cos(t) + b sin(t) = c [ \(\displaystyle \frac{a}{c}\) cos(t) + \(\displaystyle \frac{b}{c}\) sin(t) ] = c [ sin\(\displaystyle (\theta)\) cos(t) + cos\(\displaystyle (\theta)\) sin(t) ] = c sin\(\displaystyle (t + \theta)\)

 

[stapel]

Thanks to Subhotosh Khan for the correction:

Note that tan(x) = sin(x)/cos(x), so you have:

. . . . .h cos²(x) + d cos(x) = 0.5gd/v²

Wrong! In fact, you have:

. . . . .h cos²(x) + d sin(x)cos(x) = 0.5gd/v²

...which is not quite as helpful as first thought. Sorry...

 

[Some1]

Thankyou Ishuda for the tip

(I should have mentioned that h and d are the opposite and adjacent of a triangle)

The workings would take a long time to type but basically using Ishuda's tip I got to:

sin(x + j)cos(x) * sqrt(h² + d²) = 0.5dg/v²

where j = tan^-1(h/d)

so:

sqrt(h² + d²)(sin(2x + j) + h/sqrt(h² + d²)) = dg/v²

sin(2x + j) = (dg/v² - h) / sqrt(h² + d²)

x = (sin^-1((dg/v² - h) / sqrt(h² + d²)) - tan^-1(h/d)) / 2

This should be the angle at which to shoot a projectile at a target at a speed of v that is d distance away and at a height of h under a vertical acceleration of g.

Please feel free to tell me if I have made a mistake somewhere

(just realised theres a mistake but its good enough for now)