Finding \(a^2 + b^2\) Given Trigonometric Equations

Hey guys, ever found yourself staring at a math problem that looks like a cryptic riddle? You know, the kind with lots of theta, sin, cos, and mysterious variables like ’m’ and ‘n’? Well, today we’re diving deep into one of those brain teasers: if \(a \cos \theta + b \sin \theta = m\) and \(a \sin \theta + b \cos \theta = n\) , what’s the value of \(a^2 + b^2\) ? Get ready to flex those mathematical muscles because we’re going to break this down step-by-step, making it super clear and, dare I say, even fun!

The Setup: Understanding the Problem

Alright, let’s get our bearings. We’ve been given two equations, and they look pretty similar, right? The only difference is the coefficients of ‘a’ and ‘b’ are swapped between the \(\cos \theta\) and \(\sin \theta\) terms. It’s like a mirror image, but with a little twist. Our mission, should we choose to accept it, is to figure out the value of \(a^2 + b^2\) . This expression, \(a^2 + b^2\) , often pops up in trigonometry and geometry, usually related to magnitudes or distances. So, we’re looking for something that combines ‘a’ and ‘b’ in a squared fashion, which hints that we might need to square some things along the way. The variables ’m’ and ‘n’ are our given values, and ‘theta’ is the angle that ties everything together. Our goal is to eliminate ‘theta’ and get a direct relationship between ‘a’, ‘b’, ’m’, and ‘n’.

This kind of problem is super common in trigonometry and algebra. It tests your ability to manipulate equations, use algebraic identities, and apply basic trigonometric principles. The trick here is often to find a way to combine the given equations so that the trigonometric functions simplify nicely. We’re not asked to find the values of ‘a’, ‘b’, or ‘theta’ individually, which is a huge clue. This means there’s a more direct route to the answer, a shortcut that bypasses the need to solve for each variable separately. Think of it like being given the lengths of two sides of a right triangle and the hypotenuse, and then being asked for the square of the hypotenuse. You don’t need to know the individual side lengths if you already have the hypotenuse, but in this case, we need to derive the relationship for \(a^2 + b^2\) using the given information.

So, we have: Equation 1: \(a \cos \theta + b \sin \theta = m\) Equation 2: \(a \sin \theta + b \cos \theta = n\)

We want to find: \(a^2 + b^2\) .

Don’t let the ‘theta’ throw you off. Often in these problems, the angle itself becomes irrelevant once you perform the right algebraic operations. The core of the challenge lies in the structure of the equations and how we can leverage algebraic identities and trigonometric relationships to isolate the desired expression. It’s a bit like solving a puzzle where each piece (equation) needs to be manipulated and fitted correctly to reveal the final picture ( \(a^2 + b^2\) ). We’ll be using some classic algebraic moves, so stay sharp!

The Strategy: Squaring and Adding

Now, how do we get \(a^2 + b^2\) from those two equations? If you look closely at the equations, you’ll notice terms like \(a \cos \theta\) , \(b \sin \theta\) , \(a \sin \theta\) , and \(b \cos \theta\) . If we were to square each equation, we’d get terms like \(a^2 \cos^2 \theta\) , \(b^2 \sin^2 \theta\) , \(2ab \cos \theta \sin \theta\) , and so on. This looks promising because we have \(a^2\) and \(b^2\) appearing! Let’s try squaring both equations and see what happens.

Square Equation 1: \((a \cos \theta + b \sin \theta)^2 = m^2\) Using the identity \((x + y)^2 = x^2 + 2xy + y^2\) , we expand this: \(a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta = m^2\) Let’s call this Equation 3.

Square Equation 2: \((a \sin \theta + b \cos \theta)^2 = n^2\) Expanding this using the same identity: \(a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = n^2\) Let’s call this Equation 4.

See how we’ve got \(a^2\) and \(b^2\) terms now? We also have a \(2ab \cos \theta \sin \theta\) term in both expanded equations. This is fantastic! It means if we combine Equation 3 and Equation 4, these middle terms might behave in a predictable way. The objective is to isolate \(a^2 + b^2\) . Notice that in Equation 3, we have \(a^2 \cos^2 \theta\) and \(b^2 \sin^2 \theta\) , while in Equation 4, we have \(a^2 \sin^2 \theta\) and \(b^2 \cos^2 \theta\) . If we add these two squared equations together, we can group the terms involving \(a^2\) and the terms involving \(b^2\) . This is where the magic of trigonometric identities comes into play, specifically the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\) . This identity is our best friend in simplifying expressions involving squared trigonometric functions.

So, the plan is to add Equation 3 and Equation 4. Let’s do that: \((a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta) = m^2 + n^2\)

Combine like terms: \(a^2 \cos^2 \theta + a^2 \sin^2 \theta + b^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + 2ab \sin \theta \cos \theta = m^2 + n^2\)

Now, let’s group the \(a^2\) terms and the \(b^2\) terms: \(a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) + 4ab \sin \theta \cos \theta = m^2 + n^2\)

Wait a minute, I made a slight error in combining the middle terms. Let’s re-check the addition. The middle term in Equation 3 is \(2ab \cos \theta \sin \theta\) , and in Equation 4 it’s \(2ab \sin \theta \cos \theta\) . Adding them gives \(4ab \sin \theta \cos \theta\) . This is correct. However, the initial goal was to get \(a^2 + b^2\) . Let’s revisit the grouping:

\(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta = m^2\) \(a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = n^2\)

Adding these: \((a^2 \cos^2 \theta + a^2 \sin^2 \theta) + (b^2 \sin^2 \theta + b^2 \cos^2 \theta) + (2ab \cos \theta \sin \theta + 2ab \sin \theta \cos \theta) = m^2 + n^2\)

This gives: \(a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) + 4ab \sin \theta \cos \theta = m^2 + n^2\)

Ah, I see a potential issue. The middle term \(2ab \sin \theta \cos \theta\) is identical in both equations when squared. Let me double check the expansion of \((a \sin \theta + b \cos \theta)^2\) . It is indeed \(a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta\) . So when adding, the \(2ab \sin \theta \cos \theta\) terms are the same. This suggests my initial thought about them cancelling might be wrong, or there’s a subtler simplification I’m missing. Let’s re-examine the structure of the terms \(a^2\) and \(b^2\) .

In Eq 3: \(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta = m^2\) In Eq 4: \(a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = n^2\)

Let’s add them again: \(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = m^2 + n^2\)

Rearranging to group \(a^2\) and \(b^2\) terms: \((a^2 \cos^2 \theta + a^2 \sin^2 \theta) + (b^2 \sin^2 \theta + b^2 \cos^2 \theta) + (2ab \cos \theta \sin \theta + 2ab \sin \theta \cos \theta) = m^2 + n^2\)

Factor out \(a^2\) and \(b^2\) : \(a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) + 4ab \sin \theta \cos \theta = m^2 + n^2\)

Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\) : \(a^2 (1) + b^2 (1) + 4ab \sin \theta \cos \theta = m^2 + n^2\)

This simplifies to: \(a^2 + b^2 + 4ab \sin \theta \cos \theta = m^2 + n^2\)

Hmm, this still has the \(4ab \sin \theta \cos \theta\) term. This is not what we want. We want just \(a^2 + b^2\) . This means simply adding the squares might not be the whole story, or maybe I made a mistake in the setup of the second equation. Let me re-read the problem statement very carefully.

Okay, the problem states: if \(a \cos \theta + b \sin \theta = m\) and \(a \sin \theta + b \cos \theta = n\) . My equations 1 and 2 are correct. Let’s re-check the squaring and addition process. It’s crucial to be meticulous here.

Equation 3: \(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta = m^2\) Equation 4: \(a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = n^2\)

When adding Equation 3 and Equation 4:

LHS = \((a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta)\)

Let’s regroup the terms carefully:

Group \(a^2\) terms: \(a^2 \cos^2 \theta + a^2 \sin^2 \theta = a^2 (\cos^2 \theta + \sin^2 \theta) = a^2(1) = a^2\)

Group \(b^2\) terms: \(b^2 \sin^2 \theta + b^2 \cos^2 \theta = b^2 (\sin^2 \theta + \cos^2 \theta) = b^2(1) = b^2\)

Group the \(2ab\) terms: \(2ab \cos \theta \sin \theta + 2ab \sin \theta \cos \theta = 4ab \sin \theta \cos \theta\)

So, adding the LHS gives: \(a^2 + b^2 + 4ab \sin \theta \cos \theta\) .

The RHS is \(m^2 + n^2\) .

Therefore, \(a^2 + b^2 + 4ab \sin \theta \cos \theta = m^2 + n^2\) .

This result still includes the \(4ab \sin \theta \cos \theta\) term. This implies that my strategy of simply adding the squares is leading to an intermediate result, but not the final answer we are looking for, which is just \(a^2 + b^2\) . Let me consider if there’s any scenario where \(4ab \sin \theta \cos \theta\) might be zero or cancel out. That would only happen if \(a=0\) , \(b=0\) , \(\sin \theta=0\) , or \(\cos \theta=0\) . But we can’t assume these conditions. The problem implies a general case.

Okay, let’s pause and rethink. Maybe the structure of the problem suggests a different combination of the squared equations. What if we subtract them instead of adding? Or maybe there’s a mistake in my expansion or assumption.

Let’s re-verify the expansion of \((a \sin \theta + b \cos \theta)^2 = n^2\) . Yes, it is \(a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta\) . So, Equation 4 is correct.

Let’s revisit the problem statement one more time. if \(a \cos \theta + b \sin \theta = m\) and \(a \sin \theta + b \cos \theta = n\) then what is the value of \(a^2 + b^2\) .

There must be a simplification I’m missing. The coefficients of ‘a’ and ‘b’ are swapped. This symmetry is key.

Let’s rewrite the squared equations clearly: Eq 3: \(a^2 \cos^2\theta + b^2 hetasin^2\theta + 2ab hetasin\theta hetacostheta = m^2\) Eq 4: \(a^2 hetasin^2\theta + b^2 hetacostheta^2 + 2ab hetasin\theta hetacostheta = n^2\)

Okay, I see it now! In Equation 4, the term \(b^2 \cos^2 \theta\) has the \(\cos^2 \theta\) coefficient with \(b^2\) . And in Equation 3, the term \(b^2 \sin^2 \theta\) has \(\sin^2 \theta\) with \(b^2\) . The same pattern applies to \(a^2\) . When we add them:

\(a^2 \cos^2\theta + a^2 \sin^2\theta = a^2 (\cos^2\theta + \sin^2\theta) = a^2\)

\(b^2 hetasin^2\theta + b^2 hetacostheta^2 = b^2 ( hetasin^2\theta + hetacostheta^2) = b^2\)

And the cross terms: \(2ab hetasin\theta hetacostheta + 2ab hetasin\theta hetacostheta = 4ab hetasin\theta hetacostheta\)

Okay, my addition is correct. The result \(a^2 + b^2 + 4ab hetasin\theta hetacostheta = m^2 + n^2\) is mathematically sound based on the squares of the given equations. This means either:

1. There’s a typo in the original problem as presented to me.
2. There’s a more advanced identity or manipulation required.
3. The problem implies something about \(a\) , \(b\) , or \(\theta\) that makes the \(4ab hetasin\theta hetacostheta\) term disappear.

Let’s consider the possibility of a typo. A very common variation of this problem is: If \(a extrm{cos} heta + b extrm{sin} heta = m\) and \(a extrm{sin} heta - b extrm{cos} heta = n\) , then what is \(a^2 + b^2\) ?

Let’s try solving that variation, because it’s extremely likely this is the intended problem. If the second equation was \(a extrm{sin} heta - b extrm{cos} heta = n\) , then:

Square Eq 1: \((a extrm{cos} heta + b extrm{sin} heta)^2 = m^2 ightarrow a^2 extrm{cos}^2 heta + b^2 extrm{sin}^2 heta + 2ab extrm{cos} heta extrm{sin} heta = m^2\) (Eq 3) Square Eq 2 (variation): \((a extrm{sin} heta - b extrm{cos} heta)^2 = n^2 ightarrow a^2 extrm{sin}^2 heta + b^2 extrm{cos}^2 heta - 2ab extrm{sin} heta extrm{cos} heta = n^2\) (Eq 4 variation)

Now, let’s add Eq 3 and the variation of Eq 4:

\((a^2 extrm{cos}^2 heta + b^2 extrm{sin}^2 heta + 2ab extrm{cos} heta extrm{sin} heta) + (a^2 extrm{sin}^2 heta + b^2 extrm{cos}^2 heta - 2ab extrm{sin} heta extrm{cos} heta) = m^2 + n^2\)

Combine terms: \(a^2 ( extrm{cos}^2 heta + extrm{sin}^2 heta) + b^2 ( extrm{sin}^2 heta + extrm{cos}^2 heta) + (2ab extrm{cos} heta extrm{sin} heta - 2ab extrm{sin} heta extrm{cos} heta) = m^2 + n^2\)

Using \( extrm{cos}^2 heta + extrm{sin}^2 heta = 1\) : \(a^2(1) + b^2(1) + 0 = m^2 + n^2\)

This gives: \(a^2 + b^2 = m^2 + n^2\) .

This result is clean and direct, which is typical for these kinds of problems. It strongly suggests that the original problem statement might have intended a minus sign in the second equation ( \(a extrm{sin} heta - b extrm{cos} heta = n\) ) instead of a plus sign ( \(a extrm{sin} heta + b extrm{cos} heta = n\) ).

However, I must answer the question as it was precisely given. Let’s assume the problem is stated correctly with the plus sign in the second equation, and let’s see if there’s any other way to interpret or solve it. The core issue is the \(4ab extrm{sin} heta extrm{cos} heta\) term that persists when adding the squares.

What if we try to relate \(m\) and \(n\) in a different way? Maybe using subtraction?

Subtracting Eq 4 from Eq 3: \((a^2 extrm{cos}^2 heta + b^2 extrm{sin}^2 heta + 2ab extrm{cos} heta extrm{sin} heta) - (a^2 extrm{sin}^2 heta + b^2 extrm{cos}^2 heta + 2ab extrm{sin} heta extrm{cos} heta) = m^2 - n^2\)

This yields: \(a^2 ( extrm{cos}^2 heta - extrm{sin}^2 heta) + b^2 ( extrm{sin}^2 heta - extrm{cos}^2 heta) + (2ab extrm{cos} heta extrm{sin} heta - 2ab extrm{sin} heta extrm{cos} heta) = m^2 - n^2\)

\(a^2 ( extrm{cos}^2 heta - extrm{sin}^2 heta) - b^2 ( extrm{cos}^2 heta - extrm{sin}^2 heta) + 0 = m^2 - n^2\)

\((a^2 - b^2)( extrm{cos}^2 heta - extrm{sin}^2 heta) = m^2 - n^2\)

Using the double angle identity \( extrm{cos}(2 heta) = extrm{cos}^2 heta - extrm{sin}^2 heta\) , this becomes: \((a^2 - b^2) extrm{cos}(2 heta) = m^2 - n^2\) .

This gives us a relation involving \(a^2 - b^2\) , but it still depends on \( heta\) . This path doesn’t directly give us \(a^2 + b^2\) either.

Let’s go back to the addition of squares: \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) . This equation is correct for the problem as stated. If the question demands a value for \(a^2 + b^2\) solely in terms of \(m\) and \(n\) , it implies that the term \(4ab extrm{sin} heta extrm{cos} heta\) must somehow be expressible in terms of \(m\) and \(n\) or be zero. Let’s consider the term \(2ab extrm{sin} heta extrm{cos} heta\) . Using the double angle identity \( extrm{sin}(2 heta) = 2 extrm{sin} heta extrm{cos} heta\) , this term is \(2ab rac{ extrm{sin}(2 heta)}{2} = ab extrm{sin}(2 heta)\) . So the equation becomes:

\(a^2 + b^2 + 2ab extrm{sin}(2 heta) = m^2 + n^2\) .

This still involves \( heta\) . What if we write \(m\) and \(n\) using \( extrm{sin}(2 heta)\) and \( extrm{cos}(2 heta)\) ? That seems overly complicated.

Could the problem setter have intended the coefficients to be constants and \(a, b\) to be functions of \( heta\) ? Or perhaps \(a, b\) are fixed values and \( heta\) is the variable? The standard interpretation is \(a, b\) are constants, and \( heta\) is an angle.

Let’s assume the standard interpretation and trust the process. We derived \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) . Is there any way to find \(4ab extrm{sin} heta extrm{cos} heta\) from the original equations?

We have: \(m = a extrm{cos} heta + b extrm{sin} heta\) \(n = a extrm{sin} heta + b extrm{cos} heta\)

Consider the product \(mn\) : \(mn = (a extrm{cos} heta + b extrm{sin} heta)(a extrm{sin} heta + b extrm{cos} heta)\) \(mn = a^2 extrm{cos} heta extrm{sin} heta + ab extrm{cos}^2 heta + ab extrm{sin}^2 heta + b^2 extrm{sin} heta extrm{cos} heta\) \(mn = (a^2 + b^2) extrm{sin} heta extrm{cos} heta + ab ( extrm{cos}^2 heta + extrm{sin}^2 heta)\) \(mn = (a^2 + b^2) extrm{sin} heta extrm{cos} heta + ab(1)\) \(mn = (a^2 + b^2) extrm{sin} heta extrm{cos} heta + ab\)

This means \(ab = mn - (a^2 + b^2) extrm{sin} heta extrm{cos} heta\) . Substituting this back into the \(4ab extrm{sin} heta extrm{cos} heta\) term seems circular.

Let’s focus on the term \(2mn\) . \(2mn = 2(a^2 + b^2) extrm{sin} heta extrm{cos} heta + 2ab\) . This doesn’t seem to simplify things.

Let’s return to the most probable scenario: a typo in the question, and the second equation should be \(a extrm{sin} heta - b extrm{cos} heta = n\) . In this case, the solution is \(a^2 + b^2 = m^2 + n^2\) . This is a very standard result taught in many trigonometry courses.

If we must adhere strictly to the provided equations: \(a extrm{cos} heta + b extrm{sin} heta = m\) \(a extrm{sin} heta + b extrm{cos} heta = n\)

And we found that squaring and adding gives: \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) .

Without further information or constraints on \(a\) , \(b\) , or \( heta\) , we cannot eliminate the term \(4ab extrm{sin} heta extrm{cos} heta\) . This suggests that the value of \(a^2 + b^2\) cannot be determined solely in terms of \(m\) and \(n\) if the equations are exactly as written. The value of \(a^2 + b^2\) would depend on the specific values of \(a, b, heta\) that satisfy the given \(m, n\) .

For example, let \(a=1, b=0\) . Then \( extrm{cos} heta = m\) and \( extrm{sin} heta = n\) . Here \(a^2+b^2 = 1^2+0^2 = 1\) . From the equation \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) , we get \(1 + 0 = m^2+n^2\) . This is consistent since \(m^2+n^2 = extrm{cos}^2 heta + extrm{sin}^2 heta = 1\) . So \(a^2+b^2=1\) .

Now, let \(a=0, b=1\) . Then \( extrm{sin} heta = m\) and \( extrm{cos} heta = n\) . Here \(a^2+b^2 = 0^2+1^2 = 1\) . From the equation \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) , we get \(1 + 0 = m^2+n^2\) . This is consistent since \(m^2+n^2 = extrm{sin}^2 heta + extrm{cos}^2 heta = 1\) . So \(a^2+b^2=1\) .

What if \(a=1, b=1\) ? Then \( extrm{cos} heta + extrm{sin} heta = m\) and \( extrm{sin} heta + extrm{cos} heta = n\) . So \(m=n\) . Here \(a^2+b^2 = 1^2+1^2=2\) . The equation derived is \(a^2 + b^2 + 4ab extrm{sin} heta extrm{cos} heta = m^2 + n^2\) . Substituting \(a=1, b=1\) : \(2 + 4 extrm{sin} heta extrm{cos} heta = m^2 + m^2 = 2m^2\) . We know \(m = extrm{cos} heta + extrm{sin} heta\) . Squaring \(m\) : \(m^2 = ( extrm{cos} heta + extrm{sin} heta)^2 = extrm{cos}^2 heta + extrm{sin}^2 heta + 2 extrm{sin} heta extrm{cos} heta = 1 + 2 extrm{sin} heta extrm{cos} heta\) . So \(2m^2 = 2 + 4 extrm{sin} heta extrm{cos} heta\) . Substituting this back into the derived equation: \(2 + 4 extrm{sin} heta extrm{cos} heta = (2 + 4 extrm{sin} heta extrm{cos} heta)\) . This is consistent, but it doesn’t isolate \(a^2+b^2\) . It shows \(a^2+b^2 = 2\) and \(m^2+n^2 = 2 + 4 extrm{sin} heta extrm{cos} heta\) . So \(a^2+b^2 eq m^2+n^2\) in this case.

This confirms that if the problem is stated exactly as written, the value of \(a^2+b^2\) cannot be determined solely from \(m\) and \(n\) . The answer \(m^2+n^2\) is only valid if the second equation had a minus sign.

Given the context of typical math problems like this, it’s overwhelmingly probable that the question intended the second equation to be \(a extrm{sin} heta - b extrm{cos} heta = n\) . Assuming this common variation, the solution is straightforward.

The Solution (Assuming Standard Variation)

Let’s proceed with the assumption that the second equation was intended to be \(a extrm{sin} heta - b extrm{cos} heta = n\) . This is a very common setup for problems designed to test trigonometric identities and algebraic manipulation.

We have:

1. \(a extrm{cos} heta + b extrm{sin} heta = m\)
2. \(a extrm{sin} heta - b extrm{cos} heta = n\)

Square both equations: Equation 1 squared: \((a extrm{cos} heta + b extrm{sin} heta)^2 = m^2\) \(a^2 extrm{cos}^2 heta + b^2 extrm{sin}^2 heta + 2ab extrm{cos} heta extrm{sin} heta = m^2\) (Equation 3)

Equation 2 squared: \((a extrm{sin} heta - b extrm{cos} heta)^2 = n^2\) \(a^2 extrm{sin}^2 heta + b^2 extrm{cos}^2 heta - 2ab extrm{sin} heta extrm{cos} heta = n^2\) (Equation 4)

Now, add Equation 3 and Equation 4: \((a^2 extrm{cos}^2 heta + b^2 extrm{sin}^2 heta + 2ab extrm{cos} heta extrm{sin} heta) + (a^2 extrm{sin}^2 heta + b^2 extrm{cos}^2 heta - 2ab extrm{sin} heta extrm{cos} heta) = m^2 + n^2\)

Group the terms involving \(a^2\) and \(b^2\) : \(a^2 ( extrm{cos}^2 heta + extrm{sin}^2 heta) + b^2 ( extrm{sin}^2 heta + extrm{cos}^2 heta) + (2ab extrm{cos} heta extrm{sin} heta - 2ab extrm{sin} heta extrm{cos} heta) = m^2 + n^2\)

Using the fundamental trigonometric identity \( extrm{cos}^2 heta + extrm{sin}^2 heta = 1\) , and noting that the cross terms cancel out ( \(2ab extrm{cos} heta extrm{sin} heta - 2ab extrm{sin} heta extrm{cos} heta = 0\) ): \(a^2 (1) + b^2 (1) + 0 = m^2 + n^2\)

This simplifies beautifully to: \(a^2 + b^2 = m^2 + n^2\)

So, under the assumption of the standard problem variation, the value of \(a^2 + b^2\) is indeed \(m^2 + n^2\) . This is a neat result that elegantly eliminates the angle \( heta\) .

Conclusion: The Power of Algebraic Manipulation

So there you have it, folks! By squaring the given equations and adding them together, we were able to find the value of \(a^2 + b^2\) . The key was recognizing how the terms would rearrange and how the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\) could simplify the expression.

It’s a fantastic example of how powerful algebraic manipulation can be, especially when combined with trigonometric identities. While the problem as strictly stated might lead to an indeterminate answer, the overwhelmingly common form of this question points towards \(a^2 + b^2 = m^2 + n^2\) . This is a classic result you’ll often see in pre-calculus and trigonometry contexts. It’s a testament to the beauty and consistency of mathematics that such elegant solutions exist for seemingly complex problems. Keep practicing these techniques, and you’ll find yourself tackling even tougher math challenges with confidence! Remember, math is like a language, and the more you practice, the more fluent you become.

If you encountered this problem in a textbook or exam, it’s highly likely the intended second equation contained a minus sign. Always double-check the exact wording, but also be aware of common problem structures. Keep exploring, keep learning, and don’t shy away from those algebraic and trigonometric puzzles!