Question 793819
If cos(t)=4/5, 
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cos = x/r so x = 4 and r = 5
The y = sqrt[5^2-4^2] = 3
So, sin(t) = y/4 = 3/5
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Evaluate 
(i)cos(pi-t) = cos(pi)*cos(t)+sin(pi)*sin(t)
= (-1)*(4/5) + (3/5)*0
= -4/5
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(ii)cos(t+pi) = cos(t)*cos(pi) - sin(t)*sin(pi)
= (4/5)(-1) - (3/5)(0)
= -4/5
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Cheers,
Stan H.
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