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#1 2009-01-09 05:13:26
- rishicomplex
- Member
- Registered: 2009-01-07
- Posts: 4
trigo question
how do you solve this
1/sin10-√3/cos10
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#2 2009-01-09 05:19:45
- Daniel123
- Member
- Registered: 2007-05-23
- Posts: 663
Re: trigo question
Solve? that isn't an equation.
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#3 2009-01-12 08:46:31
- mathsmypassion
- Member
- Registered: 2008-12-01
- Posts: 33
Re: trigo question
Because 1/2 = sin30 and √3/2=cos30 you can write down the expression like this:
2((sin30/sin10)-(cos30/cos10))
To be able to subtract two fraction you need same denominator so:
2((sin30×cos10-cos30×sin10)/(sin10×cos10)) same as
2(sin(30-10)/(sin20/2)) = 4sin20/sin20 = 4
I have used two trig formulae:
sina×cosb - sinb×cosa=sin(a-b) and sina×cosa = sin(2a)/2 same as 2×sina×cosa = sin(2a)
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