Thread: verifying trigonometric identities

1. verifying trigonometric identities

good evening everyone.. ^^

just wanna ask for your help with this trigonometric equation...

1 - [ sin (x) - cos ^ 2 (x) ] / sin (x) = 2 cos (x)..

how would i prove it????

thanks... thanks... thanks...

2. Hello riasantos
Originally Posted by riasantos
good evening everyone.. ^^

just wanna ask for your help with this trigonometric equation...

1 - [ sin (x) - cos ^ 2 (x) ] / sin (x) = 2 cos (x)..

how would i prove it????

thanks... thanks... thanks...
I think you must have the question wrong.
$\displaystyle 1-\frac{\sin x-\cos^2 x}{\sin x}$
$\displaystyle = 1 - \frac{\sin x}{\sin x}+\frac{\cos^2x}{\sin x}$

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

$\displaystyle = \frac{\cos^2 x}{\sin x}$
which is certainly not identically equal to $\displaystyle 2 \cos x$. Check that you have the correct expression, and get back to us.

thanks for the reply..

umm.. "1" there is a part of the numerator.. ^^

so it is like...

1-(sin x - cos^2 x)
________________ = 2cos x...
sin x

..so sorry about the error...

thanks..

4. There are two possibilities, the question is wrong or I am wrong. Could you check again?

5. I hope that you will post a reply on whether the question posted was wrong. This way, I can decide on whether I am able to help you further.

6. sorry for the late response...

that's the question written in our textbook...

i just dunno if there's any typographical error..

but if there is..

then the equation must be..

1 - ( sin x - 2 cos x)
__________________ = 2 cos x..
sin x

so..

if this is also wrong.. then i think.. we can't do anything more about it..

anyway.. thank you so much for your help.. ^^

7. $\displaystyle \frac{1-(sinx-cos^2x)}{sinx}=2cosx$

$\displaystyle LHS=\frac{1-sinx+(1-sin^2x)}{sinx}$

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

$\displaystyle =\frac{-sin^2x-sinx+2}{sinx}$

This was all I could do, sorry for being of no help.

Just to double confirm, is $\displaystyle \frac{1-(sinx-cos^2x)}{sinx}=2cosx$ the question?

8. yeah, that's it.
it's okay.. ^^

i should be the one to say sorry because i disturbed you.. ^^

so sorry.. ^^

thanks for helping... ^^

9. I am just trying to improve my skills by trying out more questions, so don't feel too bad about it, instead, you have helped me! I would gain knowledge when an expert gets in to solve this question

By the way, as I am no expert in trigonometry, don't assume anything about the question yet... The question may be set correctly but haven't found someone to conquer it yet.

10. If possible, update on your progress on this question so that all of us can benefit, be it that the question is unprovable or provable

11. We need an expert to solve this question!

"Brings UP THIS POST"

12. Originally Posted by riasantos
thanks for the reply..

umm.. "1" there is a part of the numerator.. ^^

so it is like...

1-(sin x - cos^2 x)
________________ = 2cos x...
sin x

..so sorry about the error...

thanks..
$\displaystyle \frac{1 - (\sin{x} - \cos^2{x})}{\sin{x}} = \frac{1 - [\sin{x} - (1 - \sin^2{x})]}{\sin{x}}$

$\displaystyle = \frac{1 - (\sin{x} - 1 + \sin^2{x})}{\sin{x}}$

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

$\displaystyle = -\left(\frac{\sin^2{x} + \sin{x} - 2}{\sin{x}}\right)$

$\displaystyle = - \left[\frac{(\sin{x} + 2)(\sin{x} - 1)}{\sin{x}}\right]$

$\displaystyle = \frac{(1 - \sin{x})(\sin{x} + 2)}{\sin{x}}$.

This is the furthest I can go too...

Are you sure this question is asking you to PROVE the equation (i.e. show that it is an identity) or asking you to SOLVE the equation (i.e. show the values of $\displaystyle x$ for which the equation holds true)?

13. Hello everyone
Originally Posted by Punch
We need an expert to solve this question!

"Brings UP THIS POST"
As I said in an earlier post, this is not provable. None of the suggested versions are correct either - which is why no-one has come up with a solution.

So either you must find a correct version of this question, or abandon it altogether.

14. I really need help and i need the answer to the following question with one hour

cot x + cot y = cos x sin y + sinx cos y
1 - cot x cot y sin x sin y - cos x cos y

Please i need within one hour

15. Hello bobsura1

Welcome to Math Help Forum!
Originally Posted by bobsura1
cot x + cot y = cos x sin y + sinx cos y
1 - cot x cot y sin x sin y - cos x cos y

Please i need within one hour
If you want to use Math Help Forum in future:

• Post a new question as a new thread. You are much more likely to get an early reply if you do that.
• Don't leave asking your question until the last minute.

So this answer may be too late, but here it is anyway.

Note that $\displaystyle \cot x =\frac{\cos x}{\sin x}$ and $\displaystyle \cot y=\frac{\cos y}{\sin y}$. So:
$\displaystyle \frac{\cot x + \cot y}{1-\cot x \cot y}=\frac{\dfrac{\cos x}{\sin x}+\dfrac{\cos y}{\sin y}}{1 -\dfrac{\cos x}{\sin x}\dfrac{\cos y}{\sin y}}$
Multiply 'top-and-bottom by $\displaystyle \color{red}\sin x \sin y$:
$\displaystyle =\frac{\dfrac{\cos x\color{red}\sin x \sin y}{\color{black}\sin x}+\dfrac{\cos y\color{red}\sin x \sin y}{\color{black}\sin y}}{\color{red}\sin x \sin y\color{black} -\dfrac{\cos x\cos y\color{red}\sin x \sin y}{\color{black}\sin x\sin y}}$

$\displaystyle =\frac{\cos x\sin y+\sin x \cos y}{\sin x \sin y-\cos x \cos y}$