1975 AHSME Problems/Problem 30
Contents
Problem 30
Let . Then
equals
Solution
Using the difference to product identity, we find that
is equivalent to
Since sine is an odd function, we find that
, and thus
. Using the property
, we find
We multiply the entire expression by
and use the double angle identity of sine twice to find
Using the property
, we find
Substituting this back into the equation, we have
Dividing both sides by
, we have
Solution 2
We observe that is twice of
, so evaluating
and using the double angle identify is enough to find
We first let .
Notice that
and
.
Rewriting
as
, we then use the difference of angles identify to find
Rearranged to get
Using the half-angle identify twice, and noticing that
is positive, we get
Also,
Thus we can get
Squaring both sides and simplifying, this results in
From which we get
Squaring both sides again,
Multiplying both sides by
and rearranging,
Factoring out
from the first two terms and applying difference of squares,
We know that
which is not equal to
, so we divide both sides by
Using the rational root test, we find that
is a solution of this equation. However, we also know that
is not equal to
, so we divide both sides by
to get
Using the quadratic formula, we find that
We know
must be positive, which gives us
Thus, using the double angle identify, we can find
-SaraLiu24
See Also
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