- Question #af21a - Socratic
lim_ (x->0) (1-cosx) tan^2x = 1 2 Using the trigonometric identities: 2sin^2 (x 2) = 1-cosx and: tan x = (2tan (x 2)) (1-tan^2 (x 2)) we have: (1-cosx) tan^2x= (2sin
- Please help checking proof? - Socratic
Here is my proof, which is wrong? I am proving sinx - sinxtanx =0 sinx = sinxtanx 1 = tanx Apparently I HAVE to factor out sinx Why is that? Trigonometry
- How do you prove? [cos (pi 2-x)-2cos (pi 2-x)sin^2x+cos (pi 2-x)sin^4x . . .
sinx(1 −2sin2x +sin4x)sec5(x) = tanx We can now factor the parenthesis: sinx(1 −sin2x)(1 − sin2x)sec5(x) = tanx A very familiar modified Pythagorean identity will now be implemented: 1 − sin2x = cos2x Substitute: sinx(cos2x)(cos2x)sec5(x) = tanx Simplify: sinx ⋅ cos4x ⋅ sec5x = tanx Apply the secant reciprocal identity: 1 cosx
- Question #238fc - Socratic
Given tan (cotx)=cot (tanx) => tan (cotx)=tan (pi 2-tanx) =>cotx+tanx=pi 2 =>cosx sinx+sinx cosx=pi 2 => (cos^2x+sin^2x) (sinxcosx)=pi 2 =2 (2sinxcosx)=pi 2 =>2
- Question #ccb8d - Socratic
If we factor the equation: sinx (tanx-1)=0 So: sinx=0rArrx=kpi and tanx-1=0rArrtanx=1rArrx=pi 4+kpi
- Find the c value, that proves Rolles Theorem true, if . . . - Socratic
Then, EE" a "c in (a,b)" such that "f' (c)=0 In our case, f (x)=tanx, a=0, b=pi Note that, f (x)=tanx, is not defined at x=pi 2 in [0,pi] Hence, f is not continuous on (o,pi) In other words, the condition (R_1) is violated by f Accordingly, we can not apply Rolle's Theorem to f=tan Even otherwise, AA x in RR- {2npi+-pi 2 | n in ZZ}, |secx
- If y= cos (sin X). d²y dx²+Tanx. dy dx+ycos²x=? - Socratic
Explanation: As #y=cos (sinx)# # (dy) (dx)=-sin (sinx)*cosx# i e #sin (sinx)=-1 cosx (dy) (dx)# and using product formula # (d^2y) (dx^2)=-cosxcos (sinx)cosx+sin
- Find the derivative of #1 (secx-tanx)#? - Socratic
d (dx) (1 (secx-tanx))=secxtanx+sec^2x As sec^2x=tan^2x+1, we have sec^2x-tan^2x=1 i e (secx+tanx) (secx-tanx)=1 and 1 (secx-tanx)=secx+tanx Hence d (dx) (1
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