Solve:begin mathsize 20px style space sin to the power of negative 1 end exponent x plus sin to the power of negative 1 end exponent y equals fraction numerator 2 straight pi over denominator 3 end fraction space comma space cos to the power of negative 1 end exponent x minus cos to the power of negative 1 end exponent y equals straight pi over 3 end style

Asked by sunil2791 | 21st Apr, 2017, 07:10: PM

Expert Answer:

begin mathsize 16px style sin to the power of negative 1 end exponent straight x plus sin to the power of negative 1 end exponent straight y equals fraction numerator 2 straight pi over denominator 3 end fraction space space space.... left parenthesis straight i right parenthesis
cos to the power of negative 1 end exponent straight x minus cos to the power of negative 1 end exponent straight y equals straight pi over 3 space space space.... left parenthesis ii right parenthesis
Using space sin to the power of negative 1 end exponent straight x plus cos to the power of negative 1 end exponent straight x equals straight pi over 2 comma space we space write space equation space left parenthesis ii right parenthesis space as
open parentheses straight pi over 2 minus sin to the power of negative 1 end exponent straight x close parentheses minus open parentheses straight pi over 2 minus sin to the power of negative 1 end exponent straight y close parentheses equals straight pi over 3
rightwards double arrow straight pi over 2 minus sin to the power of negative 1 end exponent straight x minus straight pi over 2 plus sin to the power of negative 1 end exponent straight y equals straight pi over 3
rightwards double arrow sin to the power of negative 1 end exponent straight y minus sin to the power of negative 1 end exponent straight x equals straight pi over 3 space space.... left parenthesis iii right parenthesis
Adding space equations space left parenthesis straight i right parenthesis space and space left parenthesis iii right parenthesis comma space we space get
2 sin to the power of negative 1 end exponent straight y equals fraction numerator 2 straight pi over denominator 3 end fraction plus straight pi over 3
rightwards double arrow 2 sin to the power of negative 1 end exponent straight y equals straight pi
rightwards double arrow sin to the power of negative 1 end exponent straight y equals straight pi over 2 rightwards double arrow straight y equals sin straight pi over 2 equals 1
Subtracting space equation space left parenthesis iii right parenthesis space from space left parenthesis straight i right parenthesis comma space we space get
2 sin to the power of negative 1 end exponent straight x equals fraction numerator 2 straight pi over denominator 3 end fraction minus straight pi over 3
rightwards double arrow 2 sin to the power of negative 1 end exponent straight x equals straight pi over 3
rightwards double arrow sin to the power of negative 1 end exponent straight x equals straight pi over 6 rightwards double arrow straight x equals sin straight pi over 6 equals 1 half
Hence comma space straight x equals 1 half space and space straight y space equals space 1.
end style

Answered by Rebecca Fernandes | 27th Nov, 2017, 12:48: PM