Concept explainers
a.
the quadrant in which u2 lies.
a.

Answer to Problem 44E
Quadrant II
Explanation of Solution
Given:
cotu=3 , π<u<3π2
Given that π<u<3π2
This implies that u lies in the Quadrant III.
So, u2 will lie in the Quadrant II
b.
To find: the exact vales of sin(u2) , cos(u2) and tan(u2) .
b.

Answer to Problem 44E
cosu2=−√10−3√1020sinu2=√10+3√1020tanu2=−3−√10
Explanation of Solution
Given:
cotu=3 , π<u<3π2
Formulas Used:
cos2x=1+cos2x2
sin2x=1−cos2x2
tanx=sinxcosx
It is known that,
cotθ=31adjopp=31
So, a right triangle can be imagined for which the side adjacent to θ is 3 units and opposite side to u is 1 unit in length. Then the hypotenuse of that triangle is given by,
Then,
hyp=√opp2+adj2=√32+12=√9+1=√10
Now since u lies in quadrant III and value of cosine is negative there, so,
cosu=adjhyp=−3√10=−3√1010
Consider,
cos2u2=1+cosu2=1+(−3√1010)2=10−3√1020
Since u2lies in quadrant III and value of cosine is negative there, so,
cosu2=−√10−3√1020
And
sin2u2=1−cosu2=1−(−3√1010)2=10+3√1020
Since u2lies in quadrant III and value of sine is positive there, so,
sinu2=√10+3√1020
And,
tanu2=sinu2cosu2=√10+3√1020−√10−3√1020=−√10+3√1010−3√10=−√10+3√1010−3√10×10+3√1010+3√10
=−√(10+3√10)2102−(3√10)2=−(10+3√10)√100−90=−(10+3√10)√10=−3−√10
Chapter 5 Solutions
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