Question 968978: if sine theta = -4/7 and is in quadrant 3, then the sum of sine theta + sine (theta-pie)+sine(2pie+theta) is?
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! if sine theta = -4/7 and is in quadrant 3, then the sum of sine theta + sine (theta-pie)+sine(2pie+theta) is?
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sinx=-4/7 (Q3)
cosx=-√(1-(sin^2(x)=-√(1-16/49)=-√(33/49)=-√33/7
sinx+sin(x-π)+sin(2π+x)
sinx+(sinxcosπ-cosxsinπ)+(sin2πcosx+cos2πsinx)
=-4/7+(-4/7*-1-(-√33/7)*0)+(0*-√33/7+1*-4/7)
=-4/7+(4/7)-4/7
=-4/7
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
if sine theta = -4/7 and is in quadrant 3, then the sum of sine theta + sine (theta-pie)+sine(2pie+theta) is?
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This problem allows and assumes much simpler solution comparing with that in the post by @lwsshak3.
Second addend,  , has the opposite value to the first addend,  .
Therefore , when added, these two addends mutually cancel each other, giving 0 (zero).
Thus the sum of the first two addends vanishes, and the given expression is reduced to the last single term
 ,
which, due to the periodicity of function sin, is simply sin(theta), which is -4/7 (given in the problem).
So, the ANSWER to the problem's question is -4/7.
Solved in a simple way, without any calculations, using the basic properties of function sin.
I am 127% sure that this is the desired and the expected method of solution to the given problem.
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