The discussion centers on the implications of squaring inequalities, specifically the expression $$-1 \le \cos\left({2x}\right) \le 1$$. Participants clarify that squaring both sides of this inequality leads to the incorrect conclusion of $$0 \le \cos^2\left({2x}\right) \le 1$$, as it neglects the fact that squaring a negative number does not yield zero. Instead, it is established that the square of any real number is non-negative, thus the correct interpretation is that $$\cos^2(2x)$$ must be greater than or equal to zero, while also being bounded above by one.
PREREQUISITESMathematics students, educators, and anyone interested in deepening their understanding of inequalities and trigonometric functions.
It doesn't. If we squared both sides and did it mechanically we'd get [math]1 \leq cos^2(2x) \leq 1[/math], which is a ridiculous statement. (It is unwise to square both sides of an inequality. Weird things happen.) Instead we set the lower bound by noting that the square of any real quantity is greater than or equal to 0.tmt said:I have
$$-1 \le \cos\left({2x}\right) \le 1 $$
If everything is squared, it goes to
$$0 \le \cos^2\left({2x}\right) \le 1 $$
and I'm not sure how $(-1)^2$ turns into $0$
In the first statement, we have a small quantity between -1 and +1.tmt said:I have: $$-1 \le \cos\left({2x}\right) \le 1 $$
If everything is squared, it goes to: $$0 \le \cos^2\left({2x}\right) \le 1 $$
and I'm not sure how $(-1)^2$ turns into $0$. . . Actually, it doesn't.