Question
If $$\cos {20^ \circ } - \sin {20^ \circ } = p$$ then $$\cos {40^ \circ }$$ is equal to
A.
$$ - p\sqrt {2 - {p^2}} $$
B.
$$ p\sqrt {2 - {p^2}} $$
C.
$$ p + \sqrt {2 - {p^2}} $$
D.
None of these
Answer :
$$ p\sqrt {2 - {p^2}} $$
Solution :
$$\eqalign{
& {\left( {\cos {{20}^ \circ } + \sin {{20}^ \circ }} \right)^2} + {\left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)^2} = 2 \cr
& \therefore \,\,\,\cos {20^ \circ } + \sin {20^ \circ } = \sqrt {2 - {p^2}} > 0 \cr
& \therefore \,\,\cos {40^ \circ } = \left( {\cos {{20}^ \circ } - \sin {{20}^ \circ }} \right)\left( {\cos {{20}^ \circ } + \sin {{20}^ \circ }} \right) = p\sqrt {2 - {p^2}} . \cr} $$