Question

Trig Identities

Answer 1

\[2\sin^2xcos^x\]

Answer 2

how would you use the double angle formula with this?

Answer 3

\[(2sinxcosx)sinx=\sin(2x)sinx\]
Something like that?

Answer 4

\[2 \sin ^{2} x\;\cos {x}\;\text{?}\]

Answer 5

this is what i am trying to understand

Answer 6

Ah. Gotcha. Right, so that was meant to be cos squared. Well, separately, the identity for reducing sin^2(x) is
\[\frac{ 1-\cos2x }{ 2 }\]
And separately, the reduction of cos^2(x) is:
\[\frac{ 1+\cos2x }{ 2 } \]
Does that first part make sense?

Answer 7

i just don't understand why they used double angle.

Answer 8

i need like a walkthrough.

Answer 9

\[2\sin^2{x}\;\cos^2{x}=\frac{ 1 }{ 2 }\left( 2\sin{x}\,\cos{x} \right)^2=\frac{ 1 }{ 2 }\left( \sin{2x}\right)^2\]

Answer 10

where did the half come from

Answer 11

the 2 inside is squared but originally you only had one 2. so i multiply by 2 but need to divide by 2 (or multiply by 1/2) to maintain the original value.
it's an old algebra trick... multiplying by 1. you probably use it with fractions.

Answer 12

why did you use that version of the double angle formula? seems a bit much.

Answer 13

my friend who took trig showed me this. but Im only familiar with the general identities

Answer 14

`i just don't understand why they used double angle.`
they used the half-angle formula for sine and cosine D:
those aren't double angle, what you have written on your paper.

Answer 15

i don't understand what you are trying to change
\[2\sin^2(x)\cos^2(x)\] in to

Answer 16

i mean what the goal is
you can rewrite it in a bunch of ways

Answer 17

\[(1-\cos4x)/4\]

Answer 18

Well, the first part she did is just what I posted, she just used a power-reducing formula to get rid of the square on sin and cos. Doing that, she had:
\[2(\frac{ 1-\cos2x }{ 2 })(\frac{ 1+\cos2x }{ 2 })\]So of course from there she multiplied it out. 2 times 2 on the bottom is obviously 4. The top part is a foil of (1-cos2x)(1+cos2x)
If you foil properly, you end up with:
\[2(\frac{ 1-\cos ^{2}2x }{ 4 })\]So of course the two got multiplied in:
\[\frac{ 2-2\cos ^{2}2x }{ 4 }\]Now from here she basically did an algebra trick to make the top look like an identity. In general:
\[2\cos ^{2}x-1=cos2x\]
The only thing missing is the we need a minus 1and no negative in front of the 2. So what she does is add 1 and subtract 1 from the top at the same time. Because this is the same as addition by 0.
\[\frac{ 2-1-2\cos ^{2}2x+1 }{ 4 }=\frac{ 1-2\cos ^{2}2x+1 }{ 4 }\]Then the cos and the plus 1 part were isolated and a negative was factored out of them, leaving us with:
\[\frac{ 1-(2\cos ^{2}2x-1) }{ 4 }=\frac{ 1-\cos4x }{ 4 }\]
I know everyone else is offering their input, I just wanted to explain what exactly was done on your paper.