Question 1033520
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If sinA=-3/5 with A in Q3 find cosA/2
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First find  cos(A) = -{{{sqrt(1-sin^2(A))}}} = -{{{sqrt(1-(-3/5)^2)}}} = -{{{sqrt(1-9/25))}}} = -{{{sqrt((25-9)/25))}}} = -{{{sqrt(16/25)}}} = {{{-4/5}}}.


   The sign is "-" before the square root, since A is in Q3.


Now,  {{{cos(A/2)}}} = -{{{sqrt((1 - cos(A))/2)}}}    (see the lesson <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A> in this site)


= {{{-sqrt((1-(-4/5))/2)}}} = {{{-sqrt(9/10)}}} = {{{-3/sqrt(10)}}} = {{{(-3*sqrt(10))/10}}}.


   The sign is "-" because {{{A/2}}} is in Q2 and cosine is negative in Q2.
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