Thread: Trig Equation with varied trig functions

1. Trig 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?

2. Originally Posted by DannyMath
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?
Are you sure you dont have these in your book like $secx = \frac{1}{cosx}$

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

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

Show your work if you have problems

3. Originally Posted by harish21
Are you sure you dont have these in your book like $secx = \frac{1}{cosx}$

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

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

Show your work if you have problems
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.

4. Originally Posted by DannyMath
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.
how about looking at the left hand side of the equation:

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

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

can you move ahead now? You should be able to get to the destination. Reply back if you still have problems

5. 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

6. Originally Posted by DannyMath
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
Are you trying to prove the given question or are you trying to find the values of x for which the equation is true?

7. Trying to find all solutions in the interval [0, 2pi).

8. Originally Posted by DannyMath
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
$\frac{(sin^2 x - (1-sin^2 x)^2)}{((1-sin^2 x)(sin x))} = sinx$

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

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

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.

9. 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!

10. Originally Posted by DannyMath
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!
[0, 2pi). thats what you said the interval of x was! And the answers you have gotten do not look correct.

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

this gives:

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

$2sin^2x = 1$

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

now find out what sinx would be? Then try to figure out the values of x from $[0, 2\pi)$ for which your result is true!

11. 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!

12. Originally Posted by DannyMath
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!
Great. I was assuming you forgot to take the square root while calculating for x. The answers look good now!

Good luck for your Chemistry final! I absolutely hate chemistry!

13. Haha the only thing I really like about Chem is the history behind it. Cheers!