Solve this trigonometry inverse functions
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Solve this trigonometry inverse functions
Question: Evaluate sin-13√2, arccos 2√2 and tan-1 (-1)
Last edited by anderson_mark on Mon Jun 24, 2013 7:06 pm; edited 1 time in total (Reason for editing : Need help)
anderson_mark- Posts : 3
Join date : 2013-06-22
Re: Solve this trigonometry inverse functions
Hello,
sin(arccos(3/5))cos(arctan(7/13)) - cos(arccos(3/5))sin(arctan(7/13))
= (4/5)(13/√(218)) - (3/5)(7/√(218))
= [52/(5√(218))] - [21 / (5√(218))]
= 31 / 5√(218)
= 31√(218) / 1090 i guess this is your answer.
http://www.paarichemresources.com/export.php
sin(arccos(3/5))cos(arctan(7/13)) - cos(arccos(3/5))sin(arctan(7/13))
= (4/5)(13/√(218)) - (3/5)(7/√(218))
= [52/(5√(218))] - [21 / (5√(218))]
= 31 / 5√(218)
= 31√(218) / 1090 i guess this is your answer.
http://www.paarichemresources.com/export.php
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