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Trig Indentity problem
- Thread starter chiurox
- Start date
chiurox said:I came across another problem:
(tanA - cotA) / (tanA + cotA)
so far for this one, I got:
sin²A - cos²A .... but is there any way I can further simplify this?
chiurox said:chiurox said:I came across another problem:
(tanA - cotA) / (tanA + cotA)
so far for this one, I got:
sin²A - cos²A .... but is there any way I can further simplify this?
ok, I used the sin²A + cos²A = 1 to substitute for the (-cos²A)
Can anyone check if the answer is -1 ??
PS: I'm still having some problems with the first question. =(
D
Deleted member 4993
Guest
chiurox said:chiurox said:chiurox said:I came across another problem:
(tanA - cotA) / (tanA + cotA)
so far for this one, I got:
sin²A - cos²A = - cos(2A)
.... but is there any way I can further simplify this?
ok, I used the sin²A + cos²A = 1 to substitute for the (-cos²A)
Can anyone check if the answer is -1 ?? ..... NO
PS: I'm still having some problems with the first question. =(
D
Deleted member 4993
Guest
chiurox said:Hi, I was wondering how to solve this:
If we establish an identity, 4sec2A - 7tan2A would equal what?
Are you saying:
4sec(2A) - 7tan(2A)
or
4sec^2(A) - 7tan^2(A) .....^2 means squared
Subhotosh Khan said:chiurox said:Hi, I was wondering how to solve this:
If we establish an identity, 4sec2A - 7tan2A would equal what?
Are you saying:
4sec(2A) - 7tan(2A)
or
4sec^2(A) - 7tan^2(A) .....^2 means squared
it's 4sec^2(A) - 7tan^2(A) sorry, I remember I typed it right...
oh, about that other question: how did you get from sin^2A - cos^A = cos2A ?