Sunday, December 17, 2017

Precalculus, Chapter 4, 4.2, Section 4.2, Problem 42

Given
cos(t) = 4/5
a) cos(pi - t)
this is of the form cos(a-b)
= cos(a)cos(b)+sin(a)sin(b)
here a= pi and b= t
so,
cos(pi - t) =cos(pi)cos(t)+sin(pi)sin(t)

= (-1)cos(t) +0 = -cos(t) = -(4/5)

b) cos(t + pi)
similar to as above we get
cos(t+pi) = cos(pi+t) = -cos(t) = (-4/5)

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