(x + 5)(x + 8)
(x + 4)(x + 8)
x2 + 5x + not really sure what that extra plus is for... this is the most you can factor it... x(x + 5)
(x + 9)(x + 4)
4
The differentiation of sin x plus cosx is cos (x)-sin(x).
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
The derivative is 1/(1 + cosx)
negative sin(x)
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.
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
x = 3pi/4
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
(tan x- 1)/ (1+tan x)
Try integration by parts, (twice, I think) with u=(x+1) and dv/dx=cosx