Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an equation. For example the law of cosines, $c^2 = a^2 + b^2 - 2ab \cos C$

So I calculate it all and I get something like this: 2500 cos 130. I calculate the cos 130, and get -0.643 Now what? I have 2500 and -0.643. Do I multiply them? Or what?

Thanks.

share|improve this question
6  
Usually in math the convention is that when an operator is missing and arguments are just juxtaposed, the implicit operator is multiplication. –  Yuval Filmus Jun 15 '11 at 18:02
2  
So, $2ab\cos C = 2\cdot a \cdot b \cos(C)$. –  lhf Jun 15 '11 at 18:06
3  
How did you get to the cosine rule without encountering this convention earlier? I'm genuinely curious. –  ShreevatsaR Jun 15 '11 at 18:13
1  
@ShreevatsaR, I never took trig and I'm taking physics. :) –  Dan the Man Jun 15 '11 at 18:28

2 Answers 2

up vote 4 down vote accepted

Usually in math the convention is that when an operator is missing and arguments are just juxtaposed, the implicit operator is multiplication - Yuval Filmus

So, $2abcosC=2⋅a⋅bcos(C)$ - Ihf

share|improve this answer
    
Teamwork! $$$$$ –  The Chaz 2.0 Jun 15 '11 at 18:22
    
oh. duh. thanks. –  Dan the Man Aug 18 '11 at 14:32

Well as everyone points out there is a $\text{multiplication}$ operator. In other words $$2ab\cos{C} = 2 \times a \times b \times \cos{C}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.