Finally got to double angles. Anyways I need to show that these are identities.

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\$\$sin(4x) = 4 sin(x) cos(x) cos(2x)\$\$

The book does some magic and gets \$\$2(2sin(x)cos(x))cos(2x)\$\$

This makes no sense to me, if I expand that I get \$\$4sin(x)cos(2x)cos(2x)\$\$ which is not equal.  Everything starts with \$\$sin(a+b)=sin acos b+cos asin b\$\$ This is an identity, it holds for all \$a\$ and \$b\$. In particular, you"re allowed to replace \$b\$ with \$a\$, so long as you do it consistently throughout, and you get \$\$sin2a=2sin acos a\$\$ Stop me if you didn"t follow this. Now we can replace \$a\$ everywhere with \$2x\$ and get \$\$sin 4x=2sin2xcos2x\$\$ Now there"s a \$sin2x\$ in that formula; we can use double-angle on it to get \$\$sin4x=2(2sin xcos x)cos2x\$\$ Now multiplication is associative, which means as long as all we"re doing is multiplication, we don"t need parentheses. On the right side, we"re multiplying 5 things: \$\$sin4x=2 imes2 imessin x imescos x imescos2x\$\$ Finally, \$2 imes2=4\$, so \$\$sin4x=4sin xcos xcos2x\$\$ Thanks for contributing an answer to fundacionfernandovillalon.comematics Stack Exchange!

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Finding \$sin 2x\$ from transforming \$sin^4 x+ cos^4 x = frac79\$ using trigonometric identities Site design / logo © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.4.14.41981