# Simultaneous equation involving cos, sin

1. Jun 23, 2016

### urazi

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Jun 23, 2016

### 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

3. Jun 23, 2016

### Ray Vickson

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.

4. Jun 23, 2016

### SammyS

Staff Emeritus
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 .

5. Jun 23, 2016

### urazi

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

6. Jun 23, 2016

### urazi

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

7. Jun 23, 2016

### urazi

yes the equation should use 1.5 rather than 15

8. Jun 23, 2016

### SammyS

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

9. Jun 23, 2016

### 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

10. Jun 23, 2016

### urazi

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

11. Jun 23, 2016

### SammyS

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

12. Jun 23, 2016

### urazi

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)

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

13. Jun 23, 2016

### SammyS

Staff Emeritus
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) ?

14. Jun 24, 2016

### Ray Vickson

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.

15. Jun 25, 2016

### urazi

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

16. Jun 25, 2016

### urazi

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

17. Jun 25, 2016

### SammyS

Staff Emeritus
@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.
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.