頻度 1-sec^2(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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shift cos(2t)

Solution

Phase:0, Vertical:0

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Shift definition

For f(x)=A·g(Bx−C)+D, where g(x) is one of the basic trig functions,
CB is phase shift
D is vertical shift

f(x)=cos(2t)
g(Bx−C)=cos(2t), B=2, C=0, D=0
Therefore, the phase shift CB is 02

The vertical shift D is 0

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domain cos(2t)

range cos(2t)

periodicity cos(2t)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

頻度 1-sec^2(x)

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