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inverse of (y-3)^2=8(x-5)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`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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prove sin(3x)+sin(7x)cos(3x)−cos(7x) =cot(2x)

Solution

True

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Solution steps

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sin(3x)+sin(7x)cos(3x)−cos(7x) =cot(2x)

Manipulating left side

sin(3x)+sin(7x)cos(3x)−cos(7x)

Rewrite using trig identities

=cos(2x)sin(2x)

Use the basic trigonometric identity: cos(x)sin(x) =cot(x)

=cot(2x)

We showed that the two sides could take the same form

⇒ True

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

inverse of (y-3)^2=8(x-5)

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