tanx=2cscx
sinx/cosx=2/sinx
sin2x/cosx=2
sin2x=2cosx
1-cos2x=2cosx
0=cos2x+2cosx-1
cosx=(-2±√(2^2+4))/2
cosx=(-2±√8)/2
cosx=(-2±2√2)/2
cosx=-1±√2
cosx=approximately -2.41 or approximately 0.41.
Since the range of the cosine function is [-1,1], only approx. 0.41 works.
So:
cosx= approx. 0.41
Need calculator now (I went as far as I could without one!)
x=approx 1.148
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
The integral of sec(x) is ln|secx+tanx| + C Since the derivative is taken to the third power, we have to consider the chain rule; the original equation must be to the fourth power, and in order for that to be canceled out, the equation must also have had a coefficient of 1/4. 2x is also subject to the chain rule. I would suggest u substitution. integral(sec(2x))^3 dx u=2x du=2dx dx=1/2du integral (sec(u))^3 *1/2 du 1/8 secxtanx + 1/8(ln|secx+tanx|^4) + C
y = ln(tan(x)) u = tanx y =ln(u) dy/du = 1/u du/dx = sec2(x) dy/dx = dy/du * du/dx = sec2(x)/tan(x)
lim (x→0) [(x - sin x)/(tan x - x)]Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equalslim (x→0) [(x - sin x)'/(tan x - x)']= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)= lim (x→0) [(sin x)/(2sin x/cos3 x)]= lim (x→0) [(sin x cos3 x)/2sin x]= lim (x→0) (cos3 x/2)= 1/2Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.
No.
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
Tan
pi radians.
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
Oh, dude, finding a free solution manual for a specific textbook like Calculus by Thomas Finney 9th edition can be like searching for a needle in a haystack. You might have better luck asking your professor or checking out online forums where students share study resources. Remember, just because it's free doesn't mean it's easy to come by!
y' = (sec(x))^2
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
Yes, that is a shifted tanX graph, just as you would shift any graft.
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
Tanx was created in 1972-10.
NO, sinxtanx=sinxsinx/cosx since tanx is sinx/cosx this is sin^2xcosx now add cosx cosx(sin^2x+1) after factoring Does this equal tanx? No, since this would require tanx to equal cosx(sin^2x+1) and it does not.