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????
..please help me...
thanks... thanks... thanks...
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????
..please help me...
thanks... thanks... thanks...
Hello riasantosI think you must have the question wrong.$\displaystyle 1-\frac{\sin x-\cos^2 x}{\sin x}$which is certainly not identically equal to $\displaystyle 2 \cos x$. Check that you have the correct expression, and get back to us.$\displaystyle = 1 - \frac{\sin x}{\sin x}+\frac{\cos^2x}{\sin x}$
$\displaystyle = 1 - 1 + \frac{\cos^2 x}{\sin x}$
$\displaystyle = \frac{\cos^2 x}{\sin x}$
Grandad
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..
how about that???
if this is also wrong.. then i think.. we can't do anything more about it..
anyway.. thank you so much for your help.. ^^
$\displaystyle \frac{1-(sinx-cos^2x)}{sinx}=2cosx$
$\displaystyle LHS=\frac{1-sinx+(1-sin^2x)}{sinx}$
$\displaystyle =\frac{1-sinx+1-sin^2x}{sinx}$
$\displaystyle =\frac{-sin^2x-sinx+2}{sinx}$
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?
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.
$\displaystyle \frac{1 - (\sin{x} - \cos^2{x})}{\sin{x}} = \frac{1 - [\sin{x} - (1 - \sin^2{x})]}{\sin{x}}$
$\displaystyle = \frac{1 - (\sin{x} - 1 + \sin^2{x})}{\sin{x}}$
$\displaystyle = \frac{2 - \sin{x} - \sin^2{x}}{\sin{x}}$
$\displaystyle = -\left(\frac{\sin^2{x} + \sin{x} - 2}{\sin{x}}\right)$
$\displaystyle = - \left[\frac{(\sin{x} + 2)(\sin{x} - 1)}{\sin{x}}\right]$
$\displaystyle = \frac{(1 - \sin{x})(\sin{x} + 2)}{\sin{x}}$.
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)?
Hello bobsura1
Welcome to Math Help Forum!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 $\displaystyle \color{red}\sin x \sin y$:
Grandad$\displaystyle =\frac{\dfrac{\cos x\color{red}\sin x \sin y}{\color{black}\sin x}+\dfrac{\cos y\color{red}\sin x \sin y}{\color{black}\sin y}}{\color{red}\sin x \sin y\color{black} -\dfrac{\cos x\cos y\color{red}\sin x \sin y}{\color{black}\sin x\sin y}}$
$\displaystyle =\frac{\cos x\sin y+\sin x \cos y}{\sin x \sin y-\cos x \cos y}$