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by bluewater » Thu Nov 05, 2009 7:20 pm
here's the work i got so far. i guess what im trying to get help for this problem is the last few steps at the end
1. tan15 degrees
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
tan(45-30)=(tan45-tan30)/(1+tan45tan30)
tan(45-30)=[1-(√3/3)]/[(1+1(√3/3)]
tan(45-30)=[1-(√3/3)]/[(1+1(√3/3)]*3
tan(45-30)=(3-√3)/(3+√3)
answer = 2 - √3
completely stuck on this problem.
2. sin (17pi/12)
Thanks in advance.
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bluewater
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by Subhotosh Khan » Fri Nov 06, 2009 12:11 pm
bluewater wrote:here's the work i got so far. i guess what im trying to get help for this problem is the last few steps at the end
1. tan15 degrees
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
tan(45-30)=(tan45-tan30)/(1+tan45tan30)
tan(45-30)=[1-(√3/3)]/[(1+1(√3/3)]
tan(45-30)=[1-(√3/3)]/[(1+1(√3/3)]*3
tan(45-30)= (3-√3)/(3+√3) ................rationalize the denominator
tan 15° = [(3-√3)/(3+√3)] * [(3-√3)/(3-√3)] = (3-√3)2/[32 - (√3)2] = (9 - 6√3 + 3)/(9 - 3)
Now continue...
answer = 2 - √3
completely stuck on this problem.
2. sin (17pi/12) = sin(π + 5π/12) = - sin(5π/12) = - sin (π/4 + π/6) ... Now continue ...
Thanks in advance.
In mathematics, you don't understand things. You just get used to them ......John von Neumann
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Subhotosh Khan
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by Chadmama » Sat Nov 07, 2009 6:21 am
HI guys
Power-reduction formulas. 8 Product-to- sum and sum-to-product can be shown by using either the sum and difference identities or the multiple-angle formulae.and simplifying the right hand side of each formula using the Angle Sum and Difference Theorem will produce the left hand side
Thanks
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Chadmama
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by Denis » Sat Nov 07, 2009 1:34 pm
Chadmama wrote:Power-reduction formulas. 8 Product-to- sum and sum-to-product can be shown by using either the sum and difference identities or the multiple-angle formulae.and simplifying the right hand side of each formula using the Angle Sum and Difference Theorem will produce the left hand side
Thanks
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Denis
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