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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,
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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
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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?
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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
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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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