Quadratic equation and trigonometry

[Saitama]

Homework Statement

If the quadratic equation ax²+bx+c=0 has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Homework Equations

The Attempt at a Solution

Since the given quadratic has equal roots, its discriminant is zero.
$$b^2-4ac=0$$
The above is equivalent to:
$$\sin^2 B-4\sin A\sin C=0 \,\, (*)$$
Now I am thinking of writing (*) in terms of sinA and sinC. Substituting B with $\pi$-(A+C) doesn't look to me a good idea. I am honestly stuck here.

Any help is appreciated. Thanks!

 

[UltrafastPED]

Go back to your original condition from (1), and your result in (3), and eliminate the sin A * sin C in the denominator. This gives an expression in terms of sin^2 A, sin^2 B, sin^2 C ...

 

[Saitama]

Go back to your original condition from (1), and your result in (3), and eliminate the sin A * sin C in the denominator. This gives an expression in terms of sin^2 A, sin^2 B, sin^2 C ...

As per your suggestion, I get
$$4\left(\frac{\sin^2A+\sin^2C}{\sin^2B}\right)$$
How does this help?

 

[UltrafastPED]

where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Note that A,B,C are the angles of the triangle ... this provides additional constraints, and you can use any law that pertains to arbitrary triangles.

 

[Saitama]

Note that A,B,C are the angles of the triangle ... this provides additional constraints, and you can use any law that pertains to arbitrary triangles.

And which constraints am I supposed to use?

I already said that using A+B+C=$\pi$ is not a good idea. I can't think of anything else.

 

[pasmith]

Homework Statement

If the quadratic equation ax²+bx+c=0 has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Is that "the sum of those positive integers which are less than or equal to" the given quantity?

Homework Equations

The Attempt at a Solution

Since the given quadratic has equal roots, its discriminant is zero.
$$b^2-4ac=0$$
The above is equivalent to:
$$\sin^2 B-4\sin A\sin C=0 \,\, (*)$$
Now I am thinking of writing (*) in terms of sinA and sinC. Substituting B with $\pi$-(A+C) doesn't look to me a good idea. I am honestly stuck here.

Any help is appreciated. Thanks!

I would be tempted to use the law of sines to do away with trig functions entirely. Thus a \sin C = c \sin A and hence
<br /> \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} = \frac ac + \frac ca = \frac{a^2 + c^2}{ac}
Also, by the triangle inequality, a + c \geq b.

 

[Saitama]

Is that "the sum of those positive integers which are less than or equal to" the given quantity?

No, I have copied down the question word by word. Its exactly what I have written.

I would be tempted to use the law of sines to do away with trig functions entirely. Thus a \sin C = c \sin A and hence
<br /> \frac{\sin A}{\sin C}+\frac{\sin C}{\sin A} = \frac ac + \frac ca = \frac{a^2 + c^2}{ac}
Also, by the triangle inequality, a + c \geq b.

$$\frac ac +\frac ca=\frac{a^2 + c^2}{ac}=\frac{(a+c)^2-2ac}{ac}=\frac{(a+c)^2}{ac}-2$$
$$=4\left(\frac{(a+c)^2}{b^2}\right)-2$$
From the triangle inequality, a+c>=b.
Hence,
$$4\left(\frac{(a+c)^2}{b^2}\right) \geq 4 \Rightarrow 4\left(\frac{(a+c)^2}{b^2}\right)-2\geq 2$$
Anyways, the above was obvious from AM-GM inequality.

How do I determine the upper bound for the range? Just a guess, Cauchy-Schwarz?

 

[Saitama]

Can you please give me some more help? I am still stuck on this one.

 

[sankalpmittal]

Can you please give me some more help? I am still stuck on this one.

Well you can use weighted means.
If that does not help, you can use Tchebychef's inequality or Weierstress' inequality ?
I often see many books using the inequality to solve range you have to do.

However I am not an adept in using them though and its not in JEE syllabus also.

 

[Saitama]

Well you can use weighted means.
If that does not help, you can use Tchebychef's inequality or Weierstress' inequality ?
I often see many books using the inequality to solve range you have to do.

However I am not an adept in using them though and its not in JEE syllabus also.

I doubt any of these inequalities would help. Can you show how would you even apply them here?

 

[sankalpmittal]

I doubt any of these inequalities would help. Can you show how would you even apply them here?

No. Sorry just did not see the question properly.. :p

Ok, so by AM-GM inequality you already found out that the expression is greater than or equal to 2.
Hence its range is [2,∞)...

Now to find out sum of all integers, you just have to evaluate the term, I think :lim _n→∞S_n = 2+3+4+5+6+7...+...+n

Edit:

Wait the above is not making sense.

Did you make use of b^2-4ac=0 and sin^2B−4sinAsinC=0
in using AM-GM inequality ?

Also make an intuitive guesswork or find the range of sinA/sinC + sinC/sinA.. Obviously this will find additional conditions...And replace sinAsinC=sin^2B/4

Have you also tried cauchy-shwartz ?

 

[Saitama]

Ok, so by AM-GM inequality you already found out that the expression is greater than or equal to 2.
Hence its range is [2,∞)...

That isn't true. I have only proved that the lower bound for range is 2 but I still have to find the upper bound. I never stated that the upper bound is infinity.

 

[sankalpmittal]

That isn't true. I have only proved that the lower bound for range is 2 but I still have to find the upper bound. I never stated that the upper bound is infinity.

Is answer 9 ? I made two condition one from Pasmith's and other from UltrafastPED.

I also gave you hints in Edit in previous post.. I hope its 9.. :p

 

[Saitama]

Is answer 9 ? I made two condition one from Pasmith's and other from UltrafastPED.

I also gave you hints in Edit in previous post.. I hope its 9.. :p

Sankalp, please come up with a proper proof for the answer you get. 9 is incorrect.

 

[sankalpmittal]

Sankalp, please come up with a proper proof for the answer you get. 9 is incorrect.

Expression≥2 which you found using pasmith's hint.

Sorry yes 9 is incorrect. You made use of equal roots in finding Expression≥2 or not ?

You need to find the other inequality. I am already thinking of one:

If A+B+C=pi,

then

cosA + cosB + cosC ≤ 3/2

You can prove this also.

 

[Saitama]

If A+B+C=pi,

then

cosA + cosB + cosC ≤ 3/2

You can prove this also.

I honestly don't see that of any help here.

 

[Saitama]

Has anyone got some idea about this? I have made zero progress on this problem so far.

As always, help is appreciated.

 

[Saitama]

Plugging in Wolfram Alpha also gives nothing.

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=sin^2(x+y)=4sin(x)sin(y)

I hope someone can help me out with this problem.

 

[Ray Vickson]

Has anyone got some idea about this? I have made zero progress on this problem so far.

As always, help is appreciated.

Ultimately, it is your homework and you are responsible for solving the problem, or not. Sometimes the best thing to do is to say "I cannot do this problem" and just accept a less than perfect mark! I can guarantee that every one of the homework helpers here have encountered such situations themselves in the past!

 

[Saitama]

Ultimately, it is your homework and you are responsible for solving the problem, or not. Sometimes the best thing to do is to say "I cannot do this problem" and just accept a less than perfect mark! I can guarantee that every one of the homework helpers here have encountered such situations themselves in the past!

This is not a homework. Had this been a homework style problem, I would never put it here. Also, I am no longer a high school student that my work is graded. I am only solving problems and problems for a specific purpose.

Instead of a paragraph long lecture, a few hints would have been great. I very well know its my responsibility to solve the problem and most of the times, I don't put it on forums until I have spent enough time on it.

 

[Ray Vickson]

This is not a homework. Had this been a homework style problem, I would never put it here. Also, I am no longer a high school student that my work is graded. I am only solving problems and problems for a specific purpose.

Instead of a paragraph long lecture, a few hints would have been great. I very well know its my responsibility to solve the problem and most of the times, I don't put it on forums until I have spent enough time on it.

No, lots of people have given you hints, but you seem to want more and more and more. Somewhere it has to stop.

 

[Saitama]

No, lots of people have given you hints, but you seem to want more and more and more. Somewhere it has to stop.

Okay. Then tell me what you can do with those "hints".

I tried what pasmith said and yes, I was able to determine the lower bound of range. [strike]Hints[/strike]Posts by sankalp are of no use. What about the upper bound? I haven't yet got a hint for that. Even if we have reached the second page, the only useful post for me was #7 and none other so your point about me asking for more and more hints is completely wrong. Please check the replies on the previous page and tell me if you can make any use of them. Simply saying that "A,B,C are the angles of the triangle...this provides additional constraints" doesn't count as a hint. I know enough English to understand the given problem. No one is fool enough that he/she won't try any trig identities after reading the problem statement.

 

[I like Serena]

The problem statement is unclear to me as far as the sum of integers in the range mentioned is concerned.
So I will limit myself to what I can say.

I can see that we have 2 equations:
\begin{aligned}
k&=\frac{\sin A}{\sin C} + \frac{\sin C}{\sin A} \\
b^2&=4ac
\end{aligned}
The cosine rule gives us:
$$b^2 = a^2 + c^2 - 2ac \cos B$$
You already had:
$$k=\frac a c + \frac c a = \frac{a^2+c^2}{ac}$$
Combining gives us:
$$k=\frac{b^2 + 2ac \cos B}{\frac 1 4 b^2}$$
$$k=\frac{b^2 + \frac 1 2 b^2 \cos B}{\frac 1 4 b^2}$$
$$k=4+2\cos B$$
The range for this is:
$$2 \le k \le 6$$

The question remains what the possible angles for B are.
Did you try to find actual triangles ABC that satisfy the criteria?

 

[Ray Vickson]

Homework Statement

If the quadratic equation ax²+bx+c=0 has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of
$$\left(\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}\right)$$

Homework Equations

The Attempt at a Solution

Since the given quadratic has equal roots, its discriminant is zero.
$$b^2-4ac=0$$
The above is equivalent to:
$$\sin^2 B-4\sin A\sin C=0 \,\, (*)$$
Now I am thinking of writing (*) in terms of sinA and sinC. Substituting B with $\pi$-(A+C) doesn't look to me a good idea. I am honestly stuck here.

Any help is appreciated. Thanks!

I might be wrong, but I am getting that the problem is impossible; that is, there is no such nontrivial triangle. Here is what I did: I started with the re-formulation by pasmith, which uses \frac{\sin(A)}{\sin(C)} +\frac{ \sin(C)}{\sin(A)} = \frac{a^2 + c^2}{ac} = f(a,c). Presumably you want to find the smallest and largest feasible values of f(a,c), so that you can get lower and upper bounds on
S(a,c) \equiv \sum_{0 \leq n \leq f(a,c)} n.
Of course, the triangle can be scaled, because multiplying all sides by a constant k does not change f. We need some type of constraint on the size of the triangle, and we might as well choose the simplest one: ac = 1. Now we have $f(a,c) = a^2 + c^2.$ Also, we need $4ac = b^2$, so b = 2 (as -2 is not an allowed length). We also need the triangle inequality $a + c \geq 2$. The constraint set ac = 1, \; a+c \geq 2, \; a, c \geq 0
have just one feasible solution: $a = c = 1.$ That means that the triangle collapses into a line segment of length 2, so could not really be called a triangle at all.

As I said, I may be wrong, but that's my take on the problem.

 

[I like Serena]

I might be wrong, but I am getting that the problem is impossible; that is, there is no such nontrivial triangle. Here is what I did: I started with the re-formulation by pasmith, which uses \frac{\sin(A)}{\sin(C)} +\frac{ \sin(C)}{\sin(A)} = \frac{a^2 + c^2}{ac} = f(a,c). Presumably you want to find the smallest and largest feasible values of f(a,c), so that you can get lower and upper bounds on
S(a,c) \equiv \sum_{0 \leq n \leq f(a,c)} n.
Of course, the triangle can be scaled, because multiplying all sides by a constant k does not change f. We need some type of constraint on the size of the triangle, and we might as well choose the simplest one: ac = 1. Now we have $f(a,c) = a^2 + c^2.$ Also, we need $4ac = b^2$, so b = 2 (as -2 is not an allowed length). We also need the triangle inequality $a + c \leq 2$. The constraint set ac = 1, \; a+c \leq 2, \; a, c \geq 0
have just one feasible solution: $a = c = 1.$ That means that the triangle collapses into a line segment of length 2, so could not really be called a triangle at all.

As I said, I may be wrong, but that's my take on the problem.

How about (a,b,c)=(1/2, 2, 2).
Seems to me that satisfies the criteria as well.

The triangle equalities should be:
ac = 1, \; |a-c| \le b = 2 \le a+c, \; a, c \geq 0

 

[I like Serena]

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

 

[Ray Vickson]

How about (a,b,c)=(1/2, 2, 2).
Seems to me that satisfies the criteria as well.

The triangle equalities should be:
ac = 1, \; |a-c| \le b = 2 \le a+c, \; a, c \geq 0

Yes, you are right. That means that the minimization problem has $f(a,c)_{\min} = 2$ with $a = c = 1$ (collapse into a line-segment), while the maximization problem has $f(a,c)_{\max} = 6$, with $(a,c) = (1/(1+\sqrt{2}),1+\sqrt{2}).$

Basically, we can use the constraint $c = 1/a$ to express $f(a,c)$ as $F(a) = a^2 + 1/a^2$, which we want to maximize or minimize on the set $\{ a > 0, a + 1/a \geq 2 \}$.

 

[Ray Vickson]

We already found b = 2.
I don't think infinity for either a or c is an option anymore.
The triangle inequality $|a-c| \le b = 2$ prohibits it.

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

I have deleted the message you cite above.

 

[Saitama]

Thank you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.

In post #23 I stated an upper bound of 6 for f(a,c).

This upper bound of 6 is reached for:
$$a = \sqrt 2 + 1, \quad b = 2, \quad c = \sqrt 2 - 1$$

So we have $2 \le f(a,c) \le 6$ and both bounds are reachable.

From this, I interpret the problems question "the sum of integers in the range of f(a,c)" as $2+3+4+5+6=20$.

I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

 

[I like Serena]

Thanks you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.
I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

You're right. I guess a triangle should indeed have angles strictly between 0 and 180 degrees.

In the edge case A is a straight angle. That is correct.
If we exclude the edge case, angle A is slightly less than a straight angle.

 

[Saitama]

You're right. I guess a triangle should indeed have angles strictly between 0 and 180 degrees.

In the edge case A is a straight angle. That is correct.
If we exclude the edge case, angle A is slightly less than a straight angle.

So the final answer is 3+4+5=12? :)

 

[I like Serena]

Yep!

 

[Saitama]

Yep!

Thanks a lot I Like Serena!

 

[Ray Vickson]

Thank you both for the participation.

Continuing with k=4+2cosB, I don't see how both the bounds are reachable. Clearly, k=2 when B=$\pi$ but since this is a triangle, we can't have this or else it is a straight line. For the upper bound i.e k=6, it is possible when B=0. Again this is not possible in a triangle. So the range is 2<k<6. Sorry if this sounds dumb.



I found cosine of A for the triangle you state, it doesn't look possible. cosA comes out to be -1.

http://www.wolframalpha.com/input/?...(sqrt(2)-1)^2-(sqrt(2)+1)^2)/(2*(sqrt(2)-1)*2)

The minimum of f is 2, and that corresponds to a triangle that has collapsed into a line segment, so is not really a triangle at all. However, if you take some (a,c) giving, say f = 2.00001, there will be a legitimate triangle giving that value of f. For f = 2 the (positive) integers <= f are 1 and 2. For f = 2.00001 the positive integers <= f are still just 1 and 2. That is why f_min = 2 gives a legitimate lower bound on the integer sum, even though no actual triangle achieves that value of f.

 

[skj91]

I believe everyone is overlooking the fact that the formula which provides the solutions to the quadratic equation in the form of: ax² + bx + c is actually not

b²-4ac

but in fact it is:x = (-b ± √(b² - 4ac)) / 2a The '±' in this particular formula is the most significant mathematical notation you'll ever meet as that is precisely how you can derive your upper and lower bounds. First substitute into that formula your values and record the result of subtracting the portion under the root from -b, then do the same thing again, only find the result of adding the value under the root sign to your -b.

That is assuming b represents your y-coordinate, a is your x-coordinate and c is your hypotenuse, which is the standard representation used.

 

[skj91]

Also when dealing with sin and cos in equations where you are solving for real values of x and y, you can and will have results which have a negative value (i.e. -1) due to the way sine waves are graphed and the concept of the 'unit circle' where 0° -- (which is at the same location as 360° which is a full circle of 2π length with a center point at the coordinate (0,0) and starting point of the arc swept by its radius at the coordinate (1,0) -- will yield a value of -1 for x when the arc formed by the radius pivots around the the (0,0) point by 180° and lands on π, horizontally aligned with the x-axis. Inversely you will get a value of -1 for y when it swings another 90° past that to the 270° mark vertically aligned with the y-axis. You will, however, never produce a negative value for the radius (which is also always representative of a triangle's hypotenuse, when graphed).

Or you could go with the difference or sum of sines rules but that's probably more of a headache to get through than it's worth . Go with the ± operation on the quadratic equation over that crazy stuff.

And don't forget Pythagoreas, you may need to make use of his:

a² + b² = c²

theorem at some point.

(or x² + y² = h² in this case)

 

[skj91]

mods, please delete this (accidental double posting)

 

[Ray Vickson]

I believe everyone is overlooking the fact that the formula which provides the solutions to the quadratic equation in the form of: ax² + bx + c is actually not

b²-4ac

but in fact it is:


x = (-b ± √(b² - 4ac)) / 2a


The '±' in this particular formula is the most significant mathematical notation you'll ever meet as that is precisely how you can derive your upper and lower bounds. First substitute into that formula your values and record the result of subtracting the portion under the root from -b, then do the same thing again, only find the result of adding the value under the root sign to your -b.

That is assuming b represents your y-coordinate, a is your x-coordinate and c is your hypotenuse, which is the standard representation used.

Sorry: nobody overlooked anything. The question asked about a triangle whose sides a, b and c are coefficients of a quadratic equation having a root of multiplicity 2. Aside from that, the actual roots of the equation have nothing else to do with the problem.

 

[skj91]

Sorry: nobody overlooked anything. The question asked about a triangle whose sides a, b and c are coefficients of a quadratic equation having a root of multiplicity 2. Aside from that, the actual roots of the equation have nothing else to do with the problem.

Right, well... I'm not about to go into explaining how the coefficients of which you speak are directly proportionate to the slope of the line formed from (0,0) to the identified (a,b) or (x,y) point.-- That formed line, of course being = to the radius on a unit circle which = to the hypotenuse of a graphed triangle which is = to rise over run = y/x = opposite/adjacent = sin θ /cos θ = tan θ and that the Quadratic formula method of solving for x and then applying that to solve for their y pairs does in fact apply the correct ratio of y to x values since those coefficients are what identify what that ratio is.

Anyway, back to the OP, is there some reason you can't approach this from another methodology, for example through the Law of of Sines? Or maybe you need to look at it from the perspective of the unit circle where you have actual values to plug into those sin leg or angle variables. It all really depends on what kind of answer you (or your boss/instructor/whoever) is expecting. Is an answer in the form of another equivalent variable for sin, cos, or tan, enough? Or are you expected to find the actual number values of a specific size of triangle? If the latter is the case then you would have had to have been given some kind of known values of the specified triangle, either leg lengths, angles in degrees or the pi locations of one angle's sin and/or cos or tan functiions or any of the exact values of sin, cos, or tan when mapped to a unit circle... some kind of other info must be given or calculating any actual solution simply can't be done.


p.s. The formula I previously referenced has a misplaced parenthesis and should actually read as:

-b ± √(b² - (4ac)/2a)

 

[Saitama]

skj91, please have a look at the problem statement again. Your posts has nothing to do with solving the given problem.

 

[skj91]

It has everything to do with it, you simply aren't aware of trigonometric nor quadratic functions. It's called a SINE wave for a reason, you know. A graph of a triangle is in reality just a straight line dropped from endpoint of any radii of any circle with an origin of (0,0) wherever it happens to be resting in any of the 4 quadrants of a co-ordinate system. If you're comprehension of mathematical concepts is so limited that can't even recognize that there are multitudes of methods which can be used to come to to the same exact results then you're beyond my ability to help, but that doesn't mean you should just go and assume that I'm talking nonsense. It's not my fault if you don't know what I'm talking about. Again something that would have been no problem if you had provided a bit more info about your inquiry.

I can only assume then that this is really a simple geometric problem in which case you cross-multiply and all your sins cancel out leaving you with an answer of 1/1 which is simply 1 but I don't think that's what your actually being asked to do which is why I'm asking... don't you have any known values to work with?

Good luck.

 

[Saitama]

It has everything to do with it, you simply aren't aware of trigonometric nor quadratic functions. It's called a SINE wave for a reason, you know. A graph of a triangle is in reality just a straight line dropped from endpoint of any radii of any circle with an origin of (0,0) wherever it happens to be resting in any of the 4 quadrants of a co-ordinate system. If you're comprehension of mathematical concepts is so limited that can't even recognize that there are multitudes of methods which can be used to come to to the same exact results then you're beyond my ability to help, but that doesn't mean you should just go and assume that I'm talking nonsense. It's not my fault if you don't know what I'm talking about. Again something that would have been no problem if you had provided a bit more info about your inquiry.

I can only assume then that this is really a simple geometric problem in which case you cross-multiply and all your sins cancel out leaving you with an answer of 1/1 which is simply 1 but I don't think that's what your actually being asked to do which is why I'm asking... don't you have any known values to work with?

Good luck.

Did you have a look at the problem?

I am completely aware of most of the stuff, so please, think or at least have a look at the previous posts by me and the other posters.

 

[skj91]

Yes I looked at the problem and still you have not identified what you're trying to solve for or what method you are being asked to solve it with. You mention finding the upper and lower limits but that is impossible to do without having actual numbers to work with. In other words WHAT ARE YOUR GIVEN VALUES FOR ANY OF THE 6 SIDES OR ANGLES? It's far too difficult to attempt to explain solving equations with nothing but variables the whole way through so don't expect anyone to be willing to try doing that.
Both the equations you offer for starting points are not complete equations. SinA/SinC + SinC/SinA is not an equation, it's a meaningless pair of terms. What is it = to? What are you solving for? It can't be an equation unless there is an '=' sign somewhere so you're not giving us all the information you have to work with, here. You mention that it is a quadratic equation but then you reject using the actual "Quadratic Formula" method of solving it. That leaves only three other algebraic level methods as options.



1) Solve through Completing the square. To use this you must put your equation into a factored form. (Since you won't divulge your known values I can't explain how this is done).


2) Solve through Square Root method. Your Initial equation would have to be in the form of:

(ax-b)²-c²=0​

to use this method.


3) Solve by Factoring.

Sin²B = (sinA/sinC + SinC/SinA)² →
(SinA/SinC +SinC/SinA)(SinA/SinC +SinC/SinA) →
Sin²A/Sin²C + 2(SinASinC/SinCSinA) + Sin²C/Sin²A = Sin²B​

The only other way is to use the Quadratic formula which I have already pointed out that you're using only half of the formula required to use that method.

You tagged this thread as Trigonometry and Quadratic so I obviously assumed your question was at a college trigonometry level. Obviously that is not the case or my last 2 replies would have made perfect sense to you. This is basic high-school algebra... tag it as such.