1 / 2 ln | (1 + SiNx) / (1-sinx) | = ln | secx + TaNx |

1 / 2 ln | (1 + SiNx) / (1-sinx) | = ln | secx + TaNx |

1/2 ln|(1+sinx)/(1-sinx)|
=1/2ln|(1+sinx)²/cos²x|
=ln|(1+sinx)/cosx|
=ln|1/cosx+sinx/cosx|
=ln|secx+tanx|
Get the certificate

sinx(1+tanx*tan2/x) How did this step come out (anxious!) = SiNx (cosxcosx / 2 + sinxsinx / 2) / cosxcosx / 2 =(sinxcosx/2)/(cosxcosx/2)

Tan (x-x / 2) = (TaNx Tan (x / 2)) / (1 + TaNx * Tan (x / 2)) = (TaNx Tan (x / 2)) = (TaNx Tan (x / 2)) / Tan (x / 2)) = (TaNx Tan (x / 2)) / Tan (x / 2)) = (TaNx Tan (x / 2)) / Tan (x / 2)) = (TaNx Tan (x / 2)) / Tan (x / 2)) = (TaNx Tan (x / 2)) / (Tan (x-x / 2)) = (TaNx Tan (x / 2)) / Tan (x / 2)) = (tanx-x / 2)), which can be simplified by using the basic formula. I'm sure you have the wrong question, not tan2

The limit of (SiNx TaNx) / 4xx + TaNx approaching 0

If the title is Lim [(SiNx TaNx) / (4x ^ 2) + TaNx]
=lim[tanx(cosx-1)/(4x^2 )+tanx ]
=lim[(-x^2/2)*tanx/(4x^2 )+tanx ]
=lim[-tanx/8+tanx ]=0
If the title is Lim [(SiNx TaNx) / (4x ^ 2 + TaNx)]
=lim[tanx(cosx-1)/(4x^2 +tanx)]
=lim[(-x^3/2)/(4x^2 +tanx)] =0
subject
If Lim [(SiNx TaNx) / (4x ^ 2 * TaNx)]
=lim[(cosx-1)/(4x^2 )]
=lim[(-x^2/2)/(4x^2 )]=-1/8.

(radical 1-X-1) (x-sinx) / (x-tanx) SiNx x x approaches 0 and finds the limit

The limit is 2
(x-sinx), (x-tanx) can be reduced by equivalent infinitesimal
(Radix 1-X-1) * (Radix 1-x + 1) = x
So (radical 1-X-1) / SiNx and then equivalent infinitesimal = (radical 1-X-1) / x = (sign 1-x + 1)
So limt (radical 1-x + 1) substitutes x = 0 directly into = 2

Given that x belongs to (0, Wu) SiNx + cosx = 1 / 5, then how many one or two solutions TaNx equals depends on the process. Speed!

SiNx + cosx = 1 / 5 = = > cosx = 1 / 5 - sinxsin? X + cos? X = 1sin? X + (1 / 5 - SiNx) 2 = 125sin? X - 5sinx - 12 = 0 (5sinx - 4) (5sinx + 3) = 0sinx = 4 / 5 or SiNx = - 3 / 5 because 0 < x < π, sin

sinx=asiny tanx=btany sinx=asiny tanx=btany Where x is an acute angle, Verification: cosx = under root sign (A-1 / B-1)

tanx=btany
sinx/cosx=bsiny/cosy
Because SiNx = asiny
Cosy = B / a cosx
siny=1/a sinx
By (siny) ^ 2 + (cosy) ^ 2 = 1
1/a^2 (sinx)^2+b^2/a^2(cosx)^2=1
(sinx)^2+b^2(cosx)^2=a^2
1-(cosx)^2+b^2(cosx)^2=a^2
(b^2-1)(cosx)^2=a^2-1
(cosx)^2=(a^2-1)/(b^2-1)
Because x is an acute angle, cosx > 0
So the original proposition is proved

It is known that SiNx = asiny, Tana = btany, and the angle X is an acute angle. It is proved that: (cosx) ^ 2 = (a ^ 2-1) / (b ^ 2-1)

I think the second condition should be TaNx = btany instead of Tana = btany. It is proved that because siny = SiNx / A, tany = TaNx / B, cosy = bcosx / A, sin? Y + cos? Y = (sin? X + B? Cos? X) / a? 1 = (1-cos? X + B & S

Given the vector a = (1-sinx, 1), B = (1 / 2,1 + SiNx), if a / / B, then the acute angle X is equal to?

Vector a = (1-sinx, 1), B = (1 / 2,1 + SiNx),
If a / / b
(1-sinx)(1+sinx)=1*(1/2)
1-sin²x=1/2
sin²x=1/2
sinx=√2/2
x=45°

X is known to be an acute angle, and sinx:sinx/2=8 : 5, then cosx=

Because SiNx / sin (x / 2) = 8 / 5
So 2Sin (x / 2) cos (x / 2) / sin (x / 2) = 2cos (x / 2) = 8 / 5
So cos (x / 2) = 4 / 5
So cosx = 2 [cos (x / 2)] ^ 2-1 = 2 (4 / 5) ^ 2-1 = 7 / 25

The maximum value of the function y = sin (x / 2) cos (x / 2) - 1

y=sin(x/2)cos(x/2)-1=1/2*sinx-1
When x = 2K π + π / 2, SiNx = 1 is the maximum
So the maximum value of 1 / 2 * sinx-1 is: 1 / 2 * 1-1 = - 1 / 2
So the maximum value of y = sin (x / 2) cos (x / 2) - 1 is - 1 / 2