Question:
solve sin (x + 45 degrees) = (square root 2)cosx for 0 <_ x <_ 360 degrees?
Tamanna Shah
2011-11-29 17:20:24 UTC
solve sin (x + 45 degrees) = (square root 2)cosx for 0 <_ x <_ 360 degrees?
Two answers:
Steven Angulo
2011-11-29 17:45:46 UTC
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)



sin(x + 45) = sin(x)cos(45) + cos(x)sin(45) = sin(x)*sq rt(2)/2 + cos(x)sq rt(2)/2 =



= [sq rt(2)/2]*[sin(x) + cos(x)]



This equals sq rt(2)cos(x); therefore:



[sq rt(2)/2]*[sin(x) + cos(x)] = sq rt(2)cos(x)



sq rt(2) is multiplying both sides of the equation so it gets canceled out:



[sin(x) + cos(x)] / 2 = cos(x)



sin(x) + cos(x) = 2cos(x)



sin(x) = cos(x)



This is true for x = 45 degrees and x = 225 degrees (examining chart with the four quadrants).



Finally, we see if both values of x satisfy the original equation.



For x = 45 degrees:



sin (x + 45 degrees) = (square root 2)cosx



sin(45 + 45) = sq rt(2)cos(45) ----->



sin(90) = sq rt(2)*[sq rt(2) / 2]



1 = 2/2 -----> 1 = 1 (good)



Now for x = 225



sin (x + 45 degrees) = (square root 2)cosx



sin(225 + 45) = sq rt(2)cos(225)



sin(270) = sq rt(2)*[-sq rt(2)/2]



-1 = -2/2 -----> -1 = -1 (good)



The values for x are 45 degrees and 225 degrees. I hope this helps...
?
2017-01-12 17:01:36 UTC
Sin X 45


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