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# Trigonometric Identities

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sin x/1 + cos x + 1 + cos x/sin x = 2 csc x

Show that the left side equals the right side.

#1
+78719
+5

I think this is supposed to be :

sin x / (1 + cos x) + (1 + cos x)/sin x = 2 csc x

get a common denominator on the left   = sinx (1 + cos x)

[ sinx (sinx)   + ( 1 + cosx)^2 ] /  [ sinx (1 + cosx)]

[sin^2x + 1 + 2cosx + cos^2x] / [ sinx(1 + cosx)]

[sin^2x + cos^2x + 1 + 2cosx] / [ sinx(1 + cosx)]

{remember : sin^2x + cos^2x  = 1}

[ 1 + 1 + 2 cosx] / [sinx (1 + cosx)]

[2 + 2cos x]  /  [ sinx (1 + cos x) ]        factor out 2 on top

2(1 + cosx)  / [sinx ( 1 + cosx)]         "cancel "   the (1 + cos x) on top/bottom

2 / sinx  =

2 (1 / sinx)  =

2csc x         and this =  the right hand side

CPhill  Feb 15, 2016
edited by CPhill  Feb 15, 2016
Sort:

#1
+78719
+5

I think this is supposed to be :

sin x / (1 + cos x) + (1 + cos x)/sin x = 2 csc x

get a common denominator on the left   = sinx (1 + cos x)

[ sinx (sinx)   + ( 1 + cosx)^2 ] /  [ sinx (1 + cosx)]

[sin^2x + 1 + 2cosx + cos^2x] / [ sinx(1 + cosx)]

[sin^2x + cos^2x + 1 + 2cosx] / [ sinx(1 + cosx)]

{remember : sin^2x + cos^2x  = 1}

[ 1 + 1 + 2 cosx] / [sinx (1 + cosx)]

[2 + 2cos x]  /  [ sinx (1 + cos x) ]        factor out 2 on top

2(1 + cosx)  / [sinx ( 1 + cosx)]         "cancel "   the (1 + cos x) on top/bottom

2 / sinx  =

2 (1 / sinx)  =

2csc x         and this =  the right hand side

CPhill  Feb 15, 2016
edited by CPhill  Feb 15, 2016
#2
+466
+5

Yes, thank you very much! I had a factoring question that I wrote, but it didn't appear on the forum, so if you could answer this one too, I would greatly appreciate it: http://web2.0calc.com/questions/factoring_2753