using identities

[skittles_east_99]

i need help on this proof.

1 + cosx sinx
--------- + --------- = 2 cscx
sinx 1 + cosx


I have done this so far....

I multipled by the least common demonator which is (sinx)(1+cosx) and for the top line I got...

1 + 2cosx + cos^2x + sin^2x = 2 cscx

Then I substituted cos^2x for 1 - sin^2x and the sin^2x

1 + 2cosx + 1 - sin^2 + sin^2x

canceled out the sin^2x and not I am stuck with this

2 + 2cosx = 2 cscX

Any help would be greatly appreciated.

 

[skittles_east_99]

the problem looks like this

(1 + cosx) / sinx) + (sinx)/ (1 + cosx)

 

[stapel]

You have an unpaired parenthesis, but I think you mean the following:

. . . . .[1 + cos(x)] / [sin(x)] + [sin(x)] / [1 + cos(x)] = 2csx(x)

Noting that csc(x) = 1/sin(x), we get:

. . . . .[1 + cos(x)] / [sin(x)] + [sin(x)] / [1 + cos(x)] = 2 / sin(x)

Converting the left-hand side to the common denominator gives us:

. . . . .[(1 + cos(x))(1 + cos(x)) + sin(x)sin(x)] / [sin(x)(1 + cos(x))] = 2 / sin(x)

. . . . .[1 + 2cos(x) + cos²(x) + sin²(x)] / [sin(x)(1 + cos(x))] = 2 / sin(x)

. . . . .[1 + 2cos(x) + 1] / [sin(x)(1 + cos(x))] = 2 / sin(x)

Can you take it from there?

Eliz.

 

[skittles_east_99]

thank you!!!!!

 

[skittles_east_99]

I am still a little confused, do i distribute the sin through the demoninator and get

(2+ 2cos(x)) / (sin(x)(sincosx^2)

 

[stapel]

How did you convert "1 + cos(x)" into "sin(x)cos(x²)"? Why did you not factor the numerator and cancel with the common factor in the denominator?

Please reply showing all of your steps. Thank you.

Eliz.

 

[skittles_east_99]

i think that i was doing it wrong, i know i have to get sin by itself in the demonator, so do i multiply the top half by (1-cos(x))

 

[skittles_east_99]

i finally figured it out, thanks for all of your help.