Question

If y=sin(a+bx) then find yn?

Answer 1

`y_n=b^nsin(a+bx+n*pi/2), n in NN.`

Explanation:

We have, `y=sin(a+bx)`.

`:. y_1=dt/dx=cos(a+bx)*d/dx(a+bx)=bcos(a+bx)`.

We write this as,

`y_1=b^1*sin(a+bx+pi/2)...........................................(1)`.

Next, `y_2=(d^2y)/dx^2=d/dx{dy/dx}=d/dx{bcos(a+bx)}`,

`=b*{-sin(a+bx)}d/dx(a+bx)`,

`={-bsin(a+bx)}(b)=-b^2sin(a+bx)`.

We write this as,

`y_2=b^2sin(a+bx+pi)=b^2sin(a+bx+2*pi/2)............(2)`.

Similarly,

`y_3=d/dx{-b^2sin(a+bx)}=(-b^2)(cos(a+bx))(b), i.e.,`

`y_3=-b^3cos(a+bx)=b^3sin(a+bx+3pi/2)..................(3)`.

We conclude, `y_n=b^nsin(a+bx+n*pi/2), n in NN.`

Answer 2

`y_n=b^nsin((npi)/2+a+bx), n in NN`

Explanation:

Here,

`y=sin(a+bx)`

Diff. w. r. t. `x`,

`y_1=cos(a+bx)*b`

`y_1=bcos(a+bx)`

`color(blue)(y_1=bsin(pi/2+a+bx)`

Again diff.w.r.t. `x`

`y_2=bcos(pi/2+a+bx)*b`

`y_2=b^2sin(pi/2+pi/2+a+bx)`

#color(blue)(y_color(red)(2)=b^color(red)(2)sin(color(red)(2)* (pi/2)+a+bx)#

`y_3=b^2cos(2*(pi/2)+a+bx)*b`

`y_3=b^3sin(pi/2+2*(pi/2)+a+bx)`

#color(blue)(y_color(red)(3)=b^color(red)(3)sin(color(red)(3)* (pi/2)+a+bx)#

`y_4=b^3cos(3*(pi/2)+a+bx)*b`

`color(blue)(y_4=b^4sin(pi/2+3*(pi/2)+a+bx)`

`y_color(red)(4)=b^color(red)(4)sin(color(red)(4)*(pi/2)+a+bx)`

Similarly,

#color(blue)(y_color(red)(5)=b^color(red)(5)sin(color(red)(5)* (pi/2)+a+bx)#

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Proceeding as above,

#color(blue)(y_color(red)(n)=b^color(red)(n)sin(color(red)(n)* (pi/2)+a+bx), ninNN#

`i.e. y_n=b^nsin((npi)/2+a+bx), ninNN`

NOTE :
`color(violet)(costheta=sin(pi/2+theta)`
If we take,

`theta=px+q ,then`

`cos(px+q)=sin(pi/2+px+q)....etc.`