Find the transformation relationship between the initial three trigonometric functions sin Tan cos. For example, sin40 is equal to Cos 50

Find the transformation relationship between the initial three trigonometric functions sin Tan cos. For example, sin40 is equal to Cos 50

Sum and difference cos of two corners（ α+β）= cos α· cos β- sin α· sin β cos（ α-β）= cos α· cos β+ sin α· sin β sin（ α+β）= sin α· cos β+ cos α· sin β sin（ α-β）= sin α· cos β- cos α· sin β tan（ α+β）= (tan α+ tan β）/ （1-tan α...

Trigonometric function 30 ° 60 ° 45 ° sin Tan cos

α= 30°(π/6) sin α= 1/2 cos α= √3/2 t α n α= √3/3
α= 45°(π/4) sin α= √2/2 cos α= √2/2 t α n α= one
α= 60°(π/3) sin α= √3/2 cos α= 1/2 t α n α= √3

The second transformation method is used to find the indefinite integral of X / 1 + x ^ 4

Ling x ² = tan θ, 2x dx = sec ²θ d θ ∫ x/(1 + x^4) dx= ∫ x/(1 + tan ²θ) • sec ²θ d θ/ (2x)= (1/2)∫ sec ²θ/ sec ²θ d θ= (1/2) θ + C= (1/2)arctan(x ²) + C

How to calculate the indefinite integral of 1 / (x ^ 2 + 1) ^ (3 / 2) with the second transformation method?

Let x = tant, then t = arctanx, DX = (sect) ^ 2 DT
Substitute ∫ sect ^ 2 DT / (sect) ^ 3 = ∫ costdt = Sint + C = x / (1 + x ^ 2) ^ (1 / 2) + C

How to calculate the indefinite integral of 1 / (x (x ^ 7 + 2) with the second transformation method? (let x = 1 / t do it)

First use the second kind of substitution method, and then make up the differential
x=1/t dx=-1/t ² dt
Original formula = ∫ 1 / [(1 / T) * (1 / T ^ 7 + 1] * (- 1 / T) ²) dt =-∫ (t^6)/(1+t^7)dt=(-1/7)j∫1/(1+t^7) d(1+t^7)
-(1/7)ln(1+t^7) +C
=-(1/7)ln[1+(1/x)^7] +C
=-(1/7)ln(1+x^7) +lnx +C

The second transformation method is used for the indefinite integral of X-1 / 1 + x ^ 2

∫(x-1)/(1+x^2)dx
=∫x/(1+x^2)dx-∫1/(1+x^2)dx
=1/2∫1/(1+x^2)dx^2-∫1/(1+x^2)dx
=1/2ln(1+x^2)-arctanx+C