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Mathematics 29 Online
OpenStudy (anonymous):

can you check my work please i having trouble with a minus sign here, trying to proof the derivative of sec at point a,

OpenStudy (anonymous):

OpenStudy (anonymous):

@Preetha @Zarkon :D

OpenStudy (anonymous):

@jim_thompson5910 can you spare a minute friend :D

jimthompson5910 (jim_thompson5910):

I don't know how you got line 3

OpenStudy (anonymous):

hmm let me see

OpenStudy (anonymous):

hmm you mean the identity of cos a - cos b

OpenStudy (anonymous):

i used that identity

jimthompson5910 (jim_thompson5910):

oh ok, I see now

OpenStudy (anonymous):

@iambatman can you spare a minute friend

jimthompson5910 (jim_thompson5910):

so where are you stuck exactly?

OpenStudy (anonymous):

well the answer should be tan(a)*sec(a)

OpenStudy (anonymous):

no minus sign

OpenStudy (anonymous):

im dont know how to get rid of that minus sign

jimthompson5910 (jim_thompson5910):

ok let me look it over once more

OpenStudy (anonymous):

hehe ok

jimthompson5910 (jim_thompson5910):

ok the issue is how you applied the trig identity

jimthompson5910 (jim_thompson5910):

this is what you should have cos(x) - cos(y) = -2*sin[ (x+y)/2 ]*sin[ (x-y)/2 ] cos(a) - cos(x) = -2*sin[ (a+x)/2 ]*sin[ (a-x)/2 ] cos(a) - cos(a+h) = -2*sin[ (a+a+h)/2 ]*sin[ (a-(a+h))/2 ] cos(a) - cos(a+h) = -2*sin[ (a+a+h)/2 ]*sin[ (a-a-h)/2 ] cos(a) - cos(a+h) = -2*sin[ (2a+h)/2 ]*sin[ -h/2 ] cos(a) - cos(a+h) = -2*sin[ (2a+h)/2 ]*(-sin[ h/2 ]) cos(a) - cos(a+h) = 2*sin[ (2a+h)/2 ]*sin[ h/2 ]

OpenStudy (anonymous):

ohhh thank, thank my friend :DDD!!!! ,

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