Simultaneous equation involving cos, sin

[urazi]

Homework Statement

i have solve a engineering problem until this part I'm stuck

C cos 30 - 1.375 cosθ = 0 ----(1)
C sin 30 + 1.375 sin θ = 15 ---(2)

. i have 2 unknown C and θ, however the change to cos to sin or via versa make me lost because of my poor mathematic

Homework Equations

The Attempt at a Solution

i have try changing the sin θ to √(1-cos 2θ)/2 but i still did not get the right answer
i hope that i can have a guide to solve this by using equation funtion in scientific calculator or do it manually.

TQ in advance

 

[Buzz Bloom]

Hi urazi:

Use (1) to get an expression for C. Then substitute this for C in (2).
You now have an equation of the form:

A cos θ = B sin θ + 15.​

Now square both sides and use

cos² θ = 1 - sin² θ.​

You then have a quadratic equation involving the variable sin θ.

Hope this helps.

Regards,
Buzz

 

[Ray Vickson]

Homework Statement

i have solve a engineering problem until this part I'm stuck

C cos 30 - 1.375 cosθ = 0 ----(1)
C sin 30 + 1.375 sin θ = 15 ---(2)

. i have 2 unknown C and θ, however the change to cos to sin or via versa make me lost because of my poor mathematic

Homework Equations

The Attempt at a Solution

i have try changing the sin θ to √(1-cos 2θ)/2 but i still did not get the right answer
i hope that i can have a guide to solve this by using equation funtion in scientific calculator or do it manually.

TQ in advance

Are you sure that your equations are correctly written? Your equations do not have any real solutions. You can verify this fact for yourself: just use your first equation to solve for $C$ in terms of $\cos \theta$, then substitute that expression into the second equation, to get an equation involving $\sin \theta$ and $\cos \theta$ alone (no $C$). Plot the left-hand-side of that equation for $0 \leq \theta \leq 360^o$ and you will see that there are no real values of $\theta$ that can possibly work.

You could do more: as suggested in #2, substitute $\sin \theta = \sqrt{1 - \cos^2 \theta}$ or $\sin \theta = - \sqrt{1 -\cos^2 \theta}$ to obtain (two different) quadratic equations in $x = \cos \theta$. You will find that they do not have real solutions.

 

[SammyS]

Homework Statement

i have solve a engineering problem until this part I'm stuck

C cos 30 - 1.375 cosθ = 0 ----(1)
C sin 30 + 1.375 sin θ = 15 ---(2)

I have 2 unknowns, C and θ . However, the change to cos to sin or via versa make me lost because of my poor mathematics

Hello urazi. Welcome to PF !

As Ray pointed out, there is no real solution for this system of equations.

I looked into this a bit further. It looks to me that in order for there to be real solutions, the constant on the right hand side of Eq. 2, must be between approximately ±2.52 .

Perhaps there's a typo and that should be 1.5 rather than 15 .Beyond that:
Another way to solve this system is to:

1. Solve Eq. 1 for 1.375 cosθ , then square both sides.
   
2. Solve Eq. 2 for 1.375 sinθ , then square both sides.
3. Add the two resulting equations, noting that sin²θ + cos²θ = 1

This will result in a an equation that's quadratic in C .

 

[urazi]

Hi urazi:

Use (1) to get an expression for C. Then substitute this for C in (2).
You now have an equation of the form:

A cos θ = B sin θ + 15.​

Now square both sides and use

cos² θ = 1 - sin² θ.​

You then have a quadratic equation involving the variable sin θ.

Hope this helps.

Regards,
Buzz

thank you buzz,
yes, i have check again that 15 is should be 1.5 if not i will not get = 0 for the second equation
i have try and i get theta = 56 and C = 0.88 using your solution,
before this i have try and get theta 46 and C=0.69
however the ans i should get is theta = 40.9 and C =1.2

 

[urazi]

Hello urazi. Welcome to PF !

As Ray pointed out, there is no real solution for this system of equations.

I looked into this a bit further. It looks to me that in order for there to be real solutions, the constant on the right hand side of Eq. 2, must be between approximately ±2.52 .

Perhaps there's a typo and that should be 1.5 rather than 15 .Beyond that:
Another way to solve this system is to:

1. Solve Eq. 1 for 1.375 cosθ , then square both sides.
   
2. Solve Eq. 2 for 1.375 sinθ , then square both sides.
3. Add the two resulting equations, noting that sin²θ + cos²θ = 1

This will result in a an equation that's quadratic in C .

thank you, yes the equation should be 1.5 not 15 .

 

[urazi]

Are you sure that your equations are correctly written? Your equations do not have any real solutions. You can verify this fact for yourself: just use your first equation to solve for $C$ in terms of $\cos \theta$, then substitute that expression into the second equation, to get an equation involving $\sin \theta$ and $\cos \theta$ alone (no $C$). Plot the left-hand-side of that equation for $0 \leq \theta \leq 360^o$ and you will see that there are no real values of $\theta$ that can possibly work.

You could do more: as suggested in #2, substitute $\sin \theta = \sqrt{1 - \cos^2 \theta}$ or $\sin \theta = - \sqrt{1 -\cos^2 \theta}$ to obtain (two different) quadratic equations in $x = \cos \theta$. You will find that they do not have real solutions.

yes the equation should use 1.5 rather than 15

 

[SammyS]

thank you, yes the equation should be 1.5 not 15 .

So, have you been able to solve it using any of the suggestions given ?

 

[urazi]

So, have you been able to solve it using any of the suggestions given ?

Hai sammy, i have try using your way
i got theta =46 and C =1.1
however the answer i should get is theta = 40.9 and C is 1.2

TQ

 

[urazi]

Hai sammy, i have try using your way
i got theta =46 and C =1.1
however the answer i should get is theta = 40.9 and C is 1.2

TQ

opss no i 'm using your way i get C^2 = -0.718

 

[SammyS]

Hai sammy, i have try using your way
i got theta =46 and C =1.1
however the answer i should get is theta = 40.9 and C is 1.2

TQ

I get C = 1.2 . That's approximate. Actually I get C = 1.2006939...

That's one of two answers given by the quadratic formula.

 

[urazi]

I get C = 1.2 . That's approximate. Actually I get C = 1.2006939...

That's one of two answers given by the quadratic formula.

sammy

am i doing it right?...sorry for troubling you

1.891cos^2= - 0.75 c^2 ----(1)
1.891sin^2= - 0.25 c^2 +2.25------ (2)

adding 1 to 2

1.891 (1) = - 0.75 c^2 - 0.25 c^2 +2.25
C^2-0.359 =0

 

[SammyS]

sammy

am i doing it right?...sorry for troubling you

1.891cos^2= - 0.75 c^2 ----(1)
1.891sin^2= - 0.25 c^2 +2.25------ (2)

adding 1 to 2

1.891 (1) = - 0.75 c^2 - 0.25 c^2 +2.25
C^2-0.359 =0

Well, you did comment earlier that your math skills aren't too good .

(1) Can't have a negative sign. It's the result of squaring .

(2) What do you get if you square ( 1.5 - C/2) ?

 

[Ray Vickson]

sammy

am i doing it right?...sorry for troubling you

1.891cos^2= - 0.75 c^2 ----(1)
1.891sin^2= - 0.25 c^2 +2.25------ (2)

adding 1 to 2

1.891 (1) = - 0.75 c^2 - 0.25 c^2 +2.25
C^2-0.359 =0

You are doing it the hard way. Much easier: solve for C in terms of cos(θ) from the first equation, then substitute that into the second equation to get an equation of the form a*cos(θ) + b*sin(θ) = R. Here, R = 1.5, b = 1.375 and a = 1.375*tan(30°). Now just re-express a * cos(θ) + b* sin(θ) in the form A*cos(θ+k), a single trig term with amplitude A and phase k, both of which are easily computed. [If you prefer, you could write, instead, A*sin(θ+r), with the same amplitude A but a different phase r.]
There is a standard trick for finding A and k or r, and it is used very, very widely in engineering applications.

Anyway, it is now easy to deal with the equation A*cos(θ+k) = R or A*sin(θ+r) = R.

 

[urazi]

You are doing it the hard way. Much easier: solve for C in terms of cos(θ) from the first equation, then substitute that into the second equation to get an equation of the form a*cos(θ) + b*sin(θ) = R. Here, R = 1.5, b = 1.375 and a = 1.375*tan(30°). Now just re-express a * cos(θ) + b* sin(θ) in the form A*cos(θ+k), a single trig term with amplitude A and phase k, both of which are easily computed. [If you prefer, you could write, instead, A*sin(θ+r), with the same amplitude A but a different phase r.]
There is a standard trick for finding A and k or r, and it is used very, very widely in engineering applications.

Anyway, it is now easy to deal with the equation A*cos(θ+k) = R or A*sin(θ+r) = R.

thank you very much to sammy and ray. and all who respond...i have solved the problem :) :)

 

[urazi]

thank you very much to sammy and ray. and all who respond...i have solved the problem :) :)

and thank you very much for your patience to guide me...

 

[SammyS]

You are doing it the hard way. Much easier: solve for C in terms of cos(θ) from the first equation, then substitute that into the second equation to get an equation of the form a*cos(θ) + b*sin(θ) = R. Here, R = 1.5, b = 1.375 and a = 1.375*tan(30°). Now just re-express a * cos(θ) + b* sin(θ) in the form A*cos(θ+k), a single trig term with amplitude A and phase k, both of which are easily computed. [If you prefer, you could write, instead, A*sin(θ+r), with the same amplitude A but a different phase r.]
There is a standard trick for finding A and k or r, and it is used very, very widely in engineering applications.

Anyway, it is now easy to deal with the equation A*cos(θ+k) = R or A*sin(θ+r) = R.

@urazi ,

Whether some particular approach is relatively easier or more difficult depends upon many things.

As Ray said, any expression of the form, a⋅cos(θ) + b⋅sin(θ), can be expressed as purely a sine or as purely a cosine, either one with some "phase shift". I'm not sure if the following is the "trick" Ray was thinking of, but I find it's a good way to break this down.

Start with the expression: a⋅cos(θ) + b⋅sin(θ) .
Define A as $\ A=\sqrt{a^2+b^2\,}\ .\,$ Then A is the length of the hypotenuse of a right triangle with legs of length |a| and |b| .
Write our expression as $\ A\left(\frac aA⋅\cos(θ) + \frac bA⋅\sin(θ)\right) \ .$
Then define angle $\phi$ so that either $\ \sin(\phi)=\pm\frac aA \text{ and } \cos(\phi)=\frac bA\$
or $\ \cos(\phi)=\frac aA \text{ and } \sin(\phi)=\pm\frac bA\$
(Choose sings as convenient.)
The first choice for $\phi$ gives the angle addition formula for sine, the second choice for $\phi$ gives the angle addition formula for cosine.

In the first case we get $\ \sin(\phi)⋅\cos(θ) + \cos(\phi)⋅\sin(θ) = \sin(θ+\phi)\ .$ The other case works similarly for cosine.​

While the above can very useful on many occasions, in my opinion, in the case of the problem given here, I think the method I outlined in post #4 works out very simply.By The Way: You can solve this system with a minimum of algebra by using a graphing calculator. Solve each equation for C. Each gives C as a function of θ. Graph both of these (C becomes Y₁ and Y₂, θ becomes X) and use the calculator to solve for the two points of intersection.