sed delete all the words before a match

sed works linewise, that's why your try will not work.

So how to do it with sed? Define an address range, starting from the STAC line (/^STAC$/) to the end of the file ($). Those should be printed, so everything else (!) should get deleted:

sed -i '/^STAC$/,$!d' myfile

An awk variant which prints all lines after the match (including the match):

$ awk '/^STAC$/ { out=1 } out' file
STAC
asd
as

This matches the line that only contains the string STAC and sets out to a non-zero value. For each line, if out is non-zero, print it.

Use $0 == "STAC" instead of /^STAC$/ to do a string comparison instead of a regular expression match.

Slightly more obfuscated but shorter, using the boolean result of the match with the regular expression as an integer (will be 0 for a non-match, and 1 for a match):

awk 'p += /^STAC$/' file

If the result in p is non-zero, which it will be from the point where the regular expression first matches, the current line will be printed.

Use p += ($0 == "STAC") instead of p += /^STAC$/ to do a string comparison instead of a regular expression match.

Another option would be to use a scriptable editor like ed:

printf '%s\n' '1,/^STAC/-1 d' 'wq' | ed -s myfile

This prints two commands to ed:

• delete lines from 1 through (the line before the one that starts with STAC)
• write the file back to disk and quit

The -s option inhibits ed's default printing of the number of bytes read & written.

Using awk:

awk '/^STAC$/,/$ /' input

This will print all lines between STAC and anything (including the matching lines)

Or using a grep that supports the -z option (BSD grep does not):

Treat input and output data as sequences of lines, each terminated by a zero byte (the ASCII NUL character) instead of a newline.

grep -z 'STAC' input

grep -wn STAC file.txt | cut -d":" -f 1 | xargs -I % sed '1,%d' file.txt

Get line number of the word, pass it to xargs and use sed to delete.