1/sinx-sinx=cos/tan?
Answers:
The two identities that are important for this equation are:
(sin x)/(cos x) = tan x
(sin x)^2 + (cos x)^2 = 1
I found it easiest to work on the right side of the equation, and figure it in terms of sin x.
cos x / tan x
= cos x / (sin x / cos x)
= cos x * (cos x / sin x)
= (cos x)^2 / sin x
Next, the (cos x)^2 is equal to 1 - (sin x)^2, so substituting, we get:
= (1 - (sin x)^2) / sin x
Add finally, splitting the numerator,
= 1/sin x - (sin x)^2 / sin x
= 1/sin x - sin x
Which is the left side of the equation. This equation is therefore valid for all non-zero values of sin x.
I assume you mean (1/sinx) - sinx = (cosx/tanx)?
i hope you get a right answer , that's a hard one
i would go and ask jeeves
Start with sin^2(x) + cos^2(x) = 1
cos^2(x) = 1 - sin^2(x)
divide both sides by sin(x)
cos^2(x) / sin(x) = 1/sin(x) - sin(x)
cos(x) * cos(x)/sin(x) = 1/sin(x) - sin(x)
cos(x) / tan(x) = 1/sin(x) - sin(x)
{ sin^2(x) means sin(x) squared { sin(x) * sin(x) }, if you're not familiar with the symbols }
1-sin^2x/sinx
cos^2x/sinx
cosx/(sinx/cosx)
cosx/tanx
1/s - s = c/t => t/s - ts = c => 1/c - ss/c = c => 1 - ss = cc => 1= ss + cc which is true. so the answer is yes.
Ok. it is easy.
1/sinx-sinx = 1/sinx - sin(squared)x/sinx
=1-sin(squared)x/sinx
=cos(squared)/sinx
=cosx/sinx*cosx
sinx/cosx=tanx, so cosx/sinx=1/tanx
therefore, 1/tanx*cosx=cos/tan
get it?
things you need to know to understand this:
sin(squared)x+cos(squared)x=1
sinx/cosx=tanx
Convert cos(x)/tan(x) into sines and cosines and they simplify:
tan(x) = sin(x)/cos(x) cos(x)/tan(x) = cos(x)/(sin(x)/cos(x)) = cos(x)*(cos(x)/sin(x) = (cos(x))^2/sin x
Now convert 1/sin(x) - sin(x) in to one fraction:
LCD is sin(x), sin(x)/1*(sin(x)/sin(x) = sinx^2/sin(x). 1/sin(x) - sin(x)^2/sin(x) = (1 -(sin(x))^2)/sin(x).
Now use the pythagorian Identity and subtract.
(sin(x))^2 + (cos(x))^2 = 1. 1 = (sin(x))^2 + (cos(x))^2. (1 -(sin(x))^2)/sin(x) = ((sin(x))^2 + (cos(x))^2 -(sin(x))^2). (sin(x))^2 -(sin(x))^2 + (cos(x))^2 )/sin(x). (cos(x))^2 )/sin(x)
Now compare.
(cos(x))^2 )/sin(x) = (cos(x))^2 )/sin(x)
Voila it is the same.
QED
cosx/tanx
=cosx/(sinx/cosx)=cos^2x/sinx
=(1-sin^2x)/sinx
=(1/sinx)-sinx, as required
i hope that this helps
In one circle [0, 2π[
1/sinx
x ≠ π (180º), since 1/0 does not exist
With identities you have to work with either the left side or the right side, not both. Besides, you do not use =, because it is not an equation, it is an identity, and you just have to prove it. You use ⇒
1/sinx-sinx⇒
(common denominator (sinx))
(1 - sin^2x)/sinx⇒
since
sin^2x + cos^2x =1
then 1 - sin^2x = cos^2x
You substitute that in the identity
(1 - sin^2x)/sinx⇒ cos^2x/sinx
cos^2x/sinx can be also represented by
(cosx*cosx)/sinx
Now, you can do this
cos^2x/sinx ⇒ cosx*(cosx/sinx)
You know that tanx = sinx/cosx, so
tan^-1 x = cosx/sinx
You substitute that in the identity and you get to prove it
cosx*(cosx/sinx) ⇒ cosx*(1/tanx)⇒ cosx/tanx
And there you have it!!!! I hope it helps
RHS
= cosx / tanx
= cosx / (sinx/cosx)
= (cosx)^2 / sinx
= 1-(sinx)^2 / sinx
= (1/sinx) - sinx
= LHS (Proven)
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