then by the definition of the range of arccos, α ∈ [0, π]. And
by taking cos at both side, we have
We know, if |trig(α )| = |trig( β )|, then ref( α ) = ref( β).
Hence we have
since is in the 3rd quadrant, and cos in 3rd
quadrant is negative , it implies
cos( α ) < 0. If combined with the fact that α can either be in the
1st or 2nd quadrant, we deduce
that α must be in the 2nd quad. In summary α is an angle in 2nd
quadrant with reference angle
. Hence it must be
Example 6.2 Find the exact value of
. Solution : Let
or, α in 1st or 4th quadrant.
1. Reference angle
Moreover, is in the
2nd quadrant, where tan value is negative. Hence
2. α is in the 4th quad where tan value is also negative.
Example 6.3 Find the exact value of
Let arctan(−3/4) = α, then
where or, α is
either in the 1st or 4th quadrants.
We can narrow the range of α, by observing that tan( α ) < 0, which
implies α can not be in the
first quadrant but in the 4th quadrant.
sin(2arctan(−3/4)) = sin(2 ) = 2sin(α )cos( α),
we need to find out sin( α ) and cos(α ), of which we know α in
4th quadrant, and tan(α ) = −3/4.
By using the assistant right triangle in the 4th quadrant, we have
sin( α ) = −3/5,
cos(α ) = 4/5,
sin(2α ) = 2(−3/5)(4/5).
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