Finding the Solution to an Extra Credit Math Problem

  • Thread starter hmm?
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In summary, the person was asking how to solve an equation that was given as an extra credit problem on a test, and after thinking about it, concluded that arccos(x/6) is the solution.
  • #1
hmm?
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Hello,

[tex] \arcsin( \frac{\sqrt{36-x^2}}{6})= \arccos(?) [/tex]



I was wondering how one would solve this equation; it was given as an extra credit problem on a test I took today. Being the impulsive person I am, I can't wait until monday to receive the solution.
 
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  • #2
Think about the simplest possible solution. [tex]\arcsin(x)=\arccos(?)[/tex] has what solution?
 
  • #3
for some reason, I feel it would be cos^-1(x/6)...
 
  • #4
what reason?

stimmt.
 
  • #5
hmm? said:
for some reason, I feel it would be cos^-1(x/6)...

That would be correct. For the reason think about what the arcsince function is, it is the angle whose since you are inputting, then you can draw a right triangle from the information given in the problem and find what the cosine of that angle would be.
 
  • #6
well, since I know the sin of that triangle is sqrt(36-x^2)/6

Hypotenuse=6
opp=sqrt(36-x^2)

using pathagorean theorem x^2 + y^2=6^2
y^2= 36-x^2
y=/sqrt(36-x^2)

y=opposite leg
x=adjacent leg, so
I concluded that arccos(x/6).
 
  • #7
hmm? said:
well, since I know the sin of that triangle is sqrt(36-x^2)/6

Hypotenuse=6
opp=sqrt(36-x^2)

using pathagorean theorem x^2 + y^2=6^2
y^2= 36-x^2
y=/sqrt(36-x^2)

y=opposite leg
x=adjacent leg, so
I concluded that arccos(x/6).

Yep, that's pretty much it.
 
  • #8
the funny thing about this problem is I answered it completely different, which was also wrong; I guess it was spur of the moment, and I was enduring a brain fart of epic proportion.
 
  • #9
The more elegant answer has already been given, but suppose you're not thinking of that triangle and still want to solve it. You could take the sine of both sides, the LHS then gives: sqrt(36-x²)/6.

For the RHS, you start with sin(arccos(y)) but you can use the fundamental identity cos²a+sin²a = 1 to rewrite (I'll take the positive root): sin(a) = sqrt(1-cos²(a)). So:

sin(arccos(y)) = sqrt(1-cos²(arccos(y))) = sqrt(1-y²)

Now you can solve sqrt(1-y²) = sqrt(36-x²)/6, but by inspection it's allready easy to see that a solution for y = x/6.
 

FAQ: Finding the Solution to an Extra Credit Math Problem

How can I approach solving an extra credit math problem?

The first step is to carefully read and understand the problem. Then, brainstorm different strategies and techniques that you can use to solve it. You can also try breaking the problem into smaller parts or solving similar, simpler problems to gain insight. Finally, make sure to check your work and make adjustments as needed.

What if I can't solve the problem on my own?

Don't worry! It's completely normal to struggle with math problems, especially if they are extra credit. If you're stuck, don't be afraid to ask for help from a teacher, tutor, or classmates. Collaborating with others can often lead to new insights and solutions.

Can I use a calculator to solve the problem?

It depends on the instructions given by your teacher. Some extra credit math problems may specifically require the use of a calculator, while others may not. Make sure to carefully read the instructions and follow them accordingly.

How can I check if my solution is correct?

There are a few ways to check your solution. You can try plugging your answer back into the original problem to see if it works. You can also use a calculator or online tool to verify your answer. Another option is to ask a teacher or classmate to double check your work.

Is there a specific formula or method I should use to solve the problem?

Not necessarily. Different math problems may require different approaches, so it's important to be flexible and open to trying different strategies. However, if you're struggling to come up with a solution, you can always ask your teacher for guidance on which formula or method may be useful for that particular problem.

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