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Ex-4.2 problem (iv),(viii) in detail.

Asked by ADITYA MATHEW | 8th May, 2014, 07:51: AM

Expert Answer:

$P l e a s e space m e n t i o n space t h e space q u e s t i o n space n u m b e r : A s s u m i n g space t h e space q u e s t i o n space n u m b e r space a s space 1 comma space t h e space s o l u t i o n s space a r e : left parenthesis i v right parenthesis space E v a l u a t e space sin to the power of minus 1 end exponent open parentheses sin 2 close parentheses W e space k n o w space t h a t comma space sin to the power of minus 1 end exponent open parentheses sin theta close parentheses equals pi minus theta comma space i f space theta element of open square brackets straight pi over 2 comma fraction numerator 3 straight pi over denominator 2 end fraction close square brackets T h u s comma space sin c e space 2 element of open square brackets straight pi over 2 comma fraction numerator 3 straight pi over denominator 2 end fraction close square brackets comma space w e space h a v e space sin to the power of minus 1 end exponent open parentheses sin 2 close parentheses equals pi minus 2$

$open parentheses i v close parentheses cos to the power of minus 1 end exponent open parentheses cos fraction numerator 5 straight pi over denominator 4 end fraction close parentheses W e space k n o w space t h a t space cos to the power of minus 1 end exponent open parentheses cos theta close parentheses equals 2 straight pi minus straight theta comma space if space straight theta element of open square brackets straight pi comma 2 straight pi close square brackets Thus comma space since comma space fraction numerator 5 straight pi over denominator 4 end fraction element of open square brackets straight pi comma 2 straight pi close square brackets comma space we space have cos to the power of minus 1 end exponent open parentheses cos fraction numerator 5 straight pi over denominator 4 end fraction close parentheses equals 2 straight pi minus fraction numerator 5 straight pi over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 straight pi over denominator 4 end fraction$

Answered by Vimala Ramamurthy | 8th May, 2014, 10:53: AM

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