# Simultaneous equation involving cos, sin

• urazi
In summary: although now i'm still not sure where to put the cos 2θ in order to make the quadratic equation work

## 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

## 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.

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

urazi
urazi said:

## 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

## 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.

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.

urazi
urazi said:

## 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
Buzz Bloom said:
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

SammyS said:
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 .

Ray Vickson said:
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

urazi said:
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 ?

SammyS said:
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 said:
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

urazi said:
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.

SammyS said:
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

urazi said:
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) ?

urazi
urazi said:
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
Ray Vickson said:
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 said:
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...

Ray Vickson said:
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.

## 1. What is a simultaneous equation involving cos and sin?

A simultaneous equation involving cos and sin is a system of equations where both equations contain both cosine and sine functions. These types of equations are commonly seen in trigonometry and physics problems.

## 2. How do you solve a simultaneous equation involving cos and sin?

To solve a simultaneous equation involving cos and sin, you can use the substitution method or the elimination method. The substitution method involves solving one equation for either cosine or sine and then substituting that expression into the other equation. The elimination method involves manipulating one or both equations to eliminate one of the trigonometric functions.

## 3. Can a simultaneous equation involving cos and sin have multiple solutions?

Yes, a simultaneous equation involving cos and sin can have multiple solutions. This is because cosine and sine are periodic functions with infinitely many solutions. In some cases, the equations may have no solution or an infinite number of solutions.

## 4. Are there any special cases to consider when solving a simultaneous equation involving cos and sin?

Yes, there are a few special cases to consider when solving a simultaneous equation involving cos and sin. One case is when the equations are not independent, meaning one equation is a multiple of the other. This would result in an infinite number of solutions. Another case is when one or both equations have no solution, such as when the equations are inconsistent.

## 5. What applications use simultaneous equations involving cos and sin?

Simultaneous equations involving cos and sin are commonly used in physics and engineering for solving problems involving trigonometric functions. They can also be used in navigation and geometry problems, such as finding the intersection point of two lines on a coordinate plane.