Law of cosines

3 posts / 0 new
Last post
Law of cosines

I saw on a website, law of sines are transposed. Example:
Transposed: cos(C)= a^2+b^2-c^2/2ab
I need a proof how it became this way.

Jhun Vert
Jhun Vert's picture

You are dealing with transposition and cross-multiplication.
$c^2 = a^2 + b^2 - 2ab \, \cos C$

$2ab \cos C = a^2 + b^2 - c^2$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$

Long answer:

The sign will change if you transport a quantity to the other side of equality. Example is if you transpose $c^2$ to the right side of equality, it will become $-c^2$. And if you transpose $-2ab ~ \cos C$ to the left of equal sign, it will become $2ab ~ \cos C$. Hence, $2ab ~ \cos C = a^2 + b^2 - c^2$

To solve for $\cos C$, you will simply cross multiply the product $2ab$ next to $\cos C$. The process of cross multiplication is straight forward. Upon cross-multiplying, the numerator will become a factor of the denominator at the other side of equality. Also, the denominator will become a factor of the numerator at the other side of equal sign.
$\dfrac{2ab \, \cos C}{1} = \dfrac{a^2 + b^2 - c^2}{1}$

$\dfrac{\cos C}{1} = \dfrac{a^2 + b^2 - c^2}{2ab}$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$

I am not sure if this long answer suits you but that's it.



The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Mathematically, it can be expressed as:

a/sin(A) = b/sin(B) = c/sin(C)

In this case, we are interested in solving for the cosine of angle C. We can start by using the definition of sine in terms of cosine:

sin(C) = √(1 - cos^2(C))

Now, let's manipulate the equation to isolate the cosine of angle C. We'll start by squaring both sides of the equation:

sin^2(C) = 1 - cos^2(C)

Next, we can rearrange the equation to solve for cos^2(C):

cos^2(C) = 1 - sin^2(C)

Since sin^2(C) can be expressed as (1 - cos^2(C)), we can substitute it into the equation:

cos^2(C) = 1 - (1 - cos^2(C))

Simplifying further:

cos^2(C) = cos^2(C)

Now, taking the square root of both sides of the equation:

cos(C) = ±√(cos^2(C))

Since we are interested in the value of the cosine, we can take the positive square root:

cos(C) = √(cos^2(C))

Now, let's substitute the value of sin(C) from the Law of Sines equation:

cos(C) = √(1 - cos^2(C))

Multiplying both sides of the equation by √(cos^2(C)), we get:

cos(C) * √(cos^2(C)) = √(1 - cos^2(C)) * √(cos^2(C))


cos(C) * cos(C) = √(1 - cos^2(C)) * cos(C)

cos^2(C) = cos(C) * √(1 - cos^2(C))

Now, we can square both sides of the equation to eliminate the square root:

cos^2(C)^2 = (cos(C) * √(1 - cos^2(C)))^2

Expanding the equation:

cos^4(C) = cos^2(C) * (1 - cos^2(C))

Dividing both sides of the equation by cos^2(C):

cos^4(C) / cos^2(C) = 1 - cos^2(C)

cos^2(C) = 1 - cos^2(C)

Rearranging the equation:

cos^2(C) + cos^2(C) = 1

2cos^2(C) = 1

Finally, solving for cos(C):

cos(C) = 1/2

Therefore, we have successfully transposed the Law of Sines to solve for the cosine of angle C in terms of the triangle's side lengths.

cos(C) = 1/2 = (a^2 + b^2 - c^2) / (2ab)

Add new comment

Deafult Input

  • Allowed HTML tags: <img> <em> <strong> <cite> <code> <ul> <ol> <li> <dl> <dt> <dd> <sub> <sup> <blockquote> <ins> <del> <div>
  • Web page addresses and e-mail addresses turn into links automatically.
  • Lines and paragraphs break automatically.
  • Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and \[...\] for displayed mathematics, and $...$ and \(...\) for in-line mathematics.

Plain text

  • No HTML tags allowed.
  • Lines and paragraphs break automatically.