<PRE><B>Question 921612</B>
if cos theta = 1/4 in quadrant IV find the exact value of sin( theta - pi/4)
use x for theta
cosx=1/4
sinx=-&amp;#8730;(1-cos^2x)=-&amp;#8730;(1-1/16)=-&amp;#8730;15/16=-&amp;#8730;15/4
sin(x-&amp;#960;/4)=sinxcos(&amp;#960;/4)-cosxsin(&amp;#960;/4)=-&amp;#8730;15/4*&amp;#8730;2/2-1/4*&amp;#8730;2/2=-&amp;#8730;30/8-&amp;#8730;2/8=-(&amp;#8730;30+&amp;#8730;2)/8
note: in quadrant IV cos&gt;0, sin&lt;0.
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calculator check: (set calculator to radians)
cosx=1/4
x&amp;#8776;4.965 (radians)(in quadrant IV)
x-&amp;#960;/4&amp;#8776;4.1797
sin(x-&amp;#960;/4)&amp;#8776;sin(4.1797)&amp;#8776;-0.8614
exact value=-(&amp;#8730;30+&amp;#8730;2)/8&amp;#8776;-0.8614
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note to student:
Please let me know if my solution is understandable and helpful. Of all the solutions I have submitted, I receive the least number of responses from students with trig problems.
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