
Re: Proving trigonometric identities
Posted:
Jan 24, 2009 4:37 AM


On Sat, 24 Jan 2009 00:56:58 0800 (PST), Albert <albert.xtheunknown0@gmail.com> wrote:
>Back to solving equations: >Why does my booklet say 'none' to: >Problem 3. Solving the following equations for 0 degrees <= x <= 360 >degrees. >(a) ... >(b) 3cosec x + 4 = 0; >(c)... >etc... > >3cosec x = 4 >cosec x = 4 / 3 >cosec x is quotient of the hypotenuse divided by the opposite. Our >value here is negative and since the hypotenuse is always positive, >the opposite side must be negative and hence the answer will be in one >of the left quadrants (II or III).
Don't you mean a lower quadrant III or IV (assuming they are numbered anticlockwise, starting from the positive quadrant, at the top right)?
>So then there are solutions to the >equation, aren't there? and why/why not?
I'm a bit blearyeyed, but it certainly does look to me as if (b) is equivalent to sin x = 3/4, which has one solution for 180 degrees < x < 270 degrees, one solution for 270 degrees < x < 360 degrees, and no other solutions in the range indicated. (You'll need a calculator to find the approximate values of x.)
 Angus Rodgers

