The vector a * b = / A / * / B / is the condition that vectors a and B are collinear

The former can deduce that a and B are collinear, but a and B are collinear, but a and B are not collinear

Why B = λ a, then a and B are collinear

This is the definition of real multiplication vector
(1) If λ = 0, B is a zero vector, then a and B are collinear;
(2) If λ > 0, B and a are in the same direction, then a and B are collinear;
（3）λ

When x tends to 0, find the limit of Ln (1 + x) divided by sin2x, etc!

If x tends to 0, then ln (1 + x) ~ x, that is, the limit of Ln (1 + x) / X is 1
Sin2x ~ 2x, that is, the limit of sin2x / 2x is also 1
So the original formula = Linx / 2x = 1 / 2

Integral: ∫ DX / sin2x + 2sinx
Need process, thank you!

1/[sin2x+2sinx]
=1/[2sinxcosx+2sinx]
=1 / [2sinx (1 + cosx)] (the upper and lower parts are multiplied by SiNx)
=sinx/[2sinx*sinx*(1+cosx)]
therefore
∫dx/sin2x+2sinx
=1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx
=-1 / 2 ∫ 1 / [(1 - (cosx) ^ 2) (1 + cosx)] dcosx
=-1/2∫1/[(1-t^2)(1+t)]dt
=-1/2{-1/4*ln(t-1)-1/2*1/(1+t)+1/4*ln(1+t)}+C
=1/8*(ln(cosx-1)+ln(cosx-1)*cosx+2-ln(1+cosx)-ln(1+cosx)*cosx)/(1+cosx)+C

Is 2sinx equal to sin2x?

Don't wait

∫1/(sin2x+2sinx)dx
Please write the answer process, thank you!

The answer is still right & nbsp; & nbsp; just wrong & nbsp; you just need to change the 2sin2x of the first formula to 2sinx & nbsp; and the 2cosx & nbsp; of the second formula to cosx

If angle X satisfies SiNx + cosx = m, then (1 + cos2x) / (Cotx / 2-tanx / 2) is expressed by M

(sinx+cosx)²=m²
∴1+2sinxcosx=m²
∴sin2x=m²-1
(1+cos2x)/(cotx/2-tanx/2)
=2cos²x/[(cosx/2)/(sinx/2)-(sinx/2)/(cosx/2)]
=(2cos & # 178; X * SiNx / 2 * cosx / 2) / [(COS & # 178; X / 2) - (Sin & # 178; X / 2)] [numerator denominator multiplied by SiNx / 2 * cosx / 2]
=cos²x*sinx/cosx
=sinxcosx
=1/2*sin2x
=(m²-1)/2
Use the formula:
1+cos2x=2cos²x
cos²(x/2)-sin²(x/2)=cosx
sinx/2cosx/2=1/2sinx

We know that f (x) = cos (2x - π / 3) + 2Sin (x - π / 4) cos (x - π / 4) (x ∈ R)
Find: the minimum positive period of the function; the range of the function in the interval [- π / 12, π / 2]

If f (x) = cos (2x - π / 3) + sin (2x - π / 2) = (1 / 2) cos2x + (√ 3 / 2) sin2x - cos2x = (√ 3 / 2) sin2x - (1 / 2) cos2x = sin (2x - π / 6), then the minimum positive period of function f (x) is 2 π / 2 = π. When x ∈ [- π / 12, π / 2], 2x - π / 6 ∈ [- π / 3,5 π / 6], then sin (2x - π / 6

F (x) = sin ^ 4 (x) - sinxcosx + cos ^ 4, find the range of F (x). Resolve to 1-1 / 2Sin ^ 2 (x) - SiNx, get the error between [0,1]?

That is to say
y = 1-1/2*t^2 -t
=-1/2(t+1)^2 + 3/2
t∈[-1,1]
The axis of symmetry is t = - 1
So open your mouth down
The maximum is 3 / 2
The minimum is - 1 / 2
So the range is [- 1 / 2, 3 / 2]

The vector a * b = / A / * / B / is the condition that vectors a and B are collinear