# Trigonometric Subtraction

• Nov 27th 2013, 07:01 AM
nycmath
Trigonometric Subtraction
How [6(pi) - 2sin(3pi)]/5 = 1.87?

I get zero for 2sin(3pi). There is no way the answer is 1.87.

6(pi)/5 = 1.2, right?
• Nov 27th 2013, 07:07 AM
dokrbb
Re: Trigonometric Subtraction
Quote:

Originally Posted by nycmath
How [6(pi) - 2sin(3pi)]/5 = 1.87?

I get zero for 2sin(3pi). There is no way the answer is 1.87.

6(pi)/5 = 1.2, right?

calculating in radians I actually get [6*pi]/[5] = 3.77 (double as 1.87) ...
• Nov 27th 2013, 07:08 AM
HallsofIvy
Re: Trigonometric Subtraction
Quote:

Originally Posted by nycmath
How [6(pi) - 2sin(3pi)]/5 = 1.87?

I get zero for 2sin(3pi). There is no way the answer is 1.87.

6(pi)/5 = 1.2, right?

This is clearly miswritten. That "6 pi" looks like it should be in a trig function itself. And perhaps that "5" in intended to be inside the sine? I.e. 2 sin(2pi/5)?
• Nov 27th 2013, 07:19 AM
nycmath
Re: Trigonometric Subtraction
I forgot to set my calculator to radians.
• Nov 27th 2013, 09:04 AM
topsquark
Re: Trigonometric Subtraction
Quote:

Originally Posted by nycmath
I forgot to set my calculator to radians.

Given your initial problem statement that won't matter...the only trig function you evaluated you set to 0 (correctly) so radian/degree mode will have no effect. I agree with HallsofIvy: Are you sure your problem is written correctly?

-Dan
• Nov 28th 2013, 02:18 PM
nycmath
Re: Trigonometric Subtraction
Yes, the problem is written correctly.