Hi again all,

sec x tan x - cos x cot x = sin x

I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first?

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- Apr 10th 2010, 05:28 PMDannyMathTrig Equation with varied trig functions
Hi again all,

sec x tan x - cos x cot x = sin x

I've been looking at this for about 10 mins trying to figure out what first steps to take. Is there a general rule I should follow in terms of what to do first? - Apr 10th 2010, 05:47 PMharish21
Are you sure you dont have these in your book like $\displaystyle secx = \frac{1}{cosx}$

and $\displaystyle tanx = \frac{sinx}{cosx}$ and so on?

You need to look at your book to know the relation between trigonometric identities.

Show your work if you have problems - Apr 10th 2010, 05:56 PMDannyMath
I know the trig identities and I went about trying to apply some but I have trouble knowing which to use and how to continue, I will write down what I have in my book that I stopped at:

(sin x/cos^2x) - cos x (1/tanx) = sin x

from there I tried to apply pythagorean theorem for cos^2x but I just dont have enough experience to work it out further, I did keep trying and I will keep trying but for now I am reviewing my whole term in preparation for final. I really don't mean to ask for answers, I DO want to understand these things and I've done 50 exercises just today on it. - Apr 10th 2010, 06:05 PMharish21
how about looking at the left hand side of the equation:

since $\displaystyle cotx=\frac{cosx}{sinx}$, you can write

$\displaystyle \frac{sinx}{cos^{2}x} - \frac{cos^{2}x}{sinx} = ....$

can you move ahead now? You should be able to get to the destination. Reply back if you still have problems - Apr 10th 2010, 06:26 PMDannyMath
I seem to have gotten as far as:

(sin^2 x - (1-sin^2 x)^2)/((1-sin^2 x)(sin x)) = sin x

I got that from using the cos^2 x pythagorean identity and then trying to treat them like regular subtraction for fractions (use common denominator).

This is the last section of the precal course and I will admit I have trouble with trig identities - Apr 10th 2010, 07:18 PMharish21
- Apr 10th 2010, 08:00 PMDannyMath
Trying to find all solutions in the interval [0, 2pi).

- Apr 10th 2010, 08:13 PMharish21
$\displaystyle \frac{(sin^2 x - (1-sin^2 x)^2)}{((1-sin^2 x)(sin x))} = sinx$

$\displaystyle (sin^2 x - (1-2sin^2 x+sin^4x) = sinx (1-sin^2 x) (sinx)$

$\displaystyle sin^2 x - 1 + 2sin^2 x- sin^4x = sin^2x - sin^4x$

cancel the similar terms on the left side and the right side, then you can try solving for x now. Reply if you have any errors. - Apr 12th 2010, 09:49 AMDannyMath
Haha this problem seems rather convoluted to me but maybe it's because I'm new to it! My answers ended up being:

x = pi/6, 5pi/6, 7pi/6, and 11pi/6

Hope I've come to the right solution.

Thank you! - Apr 12th 2010, 10:10 AMharish21
[0, 2pi). thats what you said the interval of x was! And the answers you have gotten do not look correct.

$\displaystyle sin^2 x - 1 + 2sin^2 x- sin^4x = sin^2x - sin^4x$

this gives:

$\displaystyle -1 + 2sin^2 x = 0$

$\displaystyle 2sin^2x = 1$

$\displaystyle sin^2x = \frac{1}{2}$

now find out what sinx would be? Then try to figure out the values of x from $\displaystyle [0, 2\pi)$ for which your result is true! - Apr 12th 2010, 10:13 AMDannyMath
Oh right, I forgot to use the square root right? So then the reference angle would be pi/4 rather than pi/6? If so then I've done all my homework and can start on chemistry final hehe.

x = pi/4, 3pi/4, 5pi/4, 7pi/4

Thanks again! - Apr 12th 2010, 10:22 AMharish21
- Apr 12th 2010, 10:31 AMDannyMath
Haha the only thing I really like about Chem is the history behind it. Cheers!