Question 1138453
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Use the identity cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)


{{{cos(x+pi/3) = cos(x)*cos(pi/3) - sin(x)*sin(pi/3)}}}


{{{cos(x+pi/3) = cos(x)*(1/2) - sin(x)*(sqrt(3)/2)}}}


{{{cos(x+pi/3) = (1/2)*cos(x) - (sqrt(3)/2)*sin(x)}}}


{{{cos(x+pi/3) = (1/2)*(cos(x)-sqrt(3)*sin(x))}}}


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Use the identity sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B)


{{{sin(x-pi/6) = sin(x)cos(pi/6) - cos(x)sin(pi/6)}}}


{{{sin(x-pi/6) = sin(x)*(sqrt(3)/2) - cos(x)(1/2)}}}


{{{sin(x-pi/6) = (sqrt(3)/2)*sin(x) - (1/2)*cos(x)}}}


{{{sin(x-pi/6) = (1/2)*(sqrt(3)*sin(x)-cos(x))}}}


{{{sin(x-pi/6) = (1/2)*(-cos(x)+sqrt(3)*sin(x))}}}


{{{sin(x-pi/6) = -(1/2)*(cos(x)-sqrt(3)*sin(x))}}}


note how this result is basically the negative version of the result from the previous section above.


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{{{cos(x+pi/3)+sin(x-pi/6) = (1/2)*(cos(x)-sqrt(3)*sin(x))-(1/2)*(cos(x)-sqrt(3)*sin(x))}}}


{{{cos(x+pi/3)+sin(x-pi/6) = (1/2)*y-(1/2)*y}}} Let y = cos(x)-sqrt(3)*sin(x)


{{{cos(x+pi/3)+sin(x-pi/6) = (1/2-1/2)*y}}}


{{{cos(x+pi/3)+sin(x-pi/6) = 0*y}}}


{{{cos(x+pi/3)+sin(x-pi/6) = 0}}}


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Answer: The original expression simplifies to <font color=red size=4>0</font>


We can say that the equation {{{cos(x+pi/3)+sin(x-pi/6) = 0}}} is an identity, meaning that it is a true equation for all real numbers x.
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