I've Invented New Trig!
sos ( x ) = sin 2 ( x ) − cos 2 ( x ) cin ( x ) = cos 2 ( x ) − sin 2 ( x )
Let us define two new trigonometric identities as shown above.
For real x , find the minimum value of 2 sos ( x ) cin ( x ) .
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2 solutions
Why can't it be 0?
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Because zero is greater than -2. and we have to find the minimum value.....
Let
a
=
sin
2
x
and
b
=
cos
2
x
So
s
o
s
x
=
a
−
b
and
c
i
n
x
=
b
−
a
2
(
a
−
b
)
(
b
−
a
)
=
2
(
−
a
2
−
b
2
+
2
a
b
)
We can now assume that 2ab is equal to its lowest real solution 0 in order to find the lowest solution possible.
(
−
a
2
−
b
2
+
0
)
=
−
sin
2
x
−
cos
2
x
=
−
1
2
×
−
1
=
−
2
Note: (2ab can't be negative for both functions, a and b, are squared trigonometric functions.)
But ( a − b ) ( b − a ) = − a 2 − b 2 + 2 a b .
How can we show that the minimum value if − 2 from that?
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Because the functions in a and b are squared functions and can only result in real positive or zero answers, we have to set 2ab's value to zero, which we have the liberty to do, for we can set x equal to pi which would effectively make a=0 (sin^2(pi)=0). After this step 2ab will also be 0. We have to set it to zero as it would only increase the solution otherwise. With 2ab equalling zero, the statement above is still true, but I'm fixing it now. (I was wondering why the only thing it could be was -2 before, so thank you.)
sos ( x ) = sin 2 x − cos 2 x = − cos ( 2 x ) cin ( x ) = cos 2 x − sin 2 x = cos ( 2 x )
y = 2 sos ( x ) cin ( x ) y = − 2 ( cos ( 2 x ) ) 2 y min = − 2