Question

Question #12b6c

Answer 1

See the proof below

Explanation:

We use

`sin^2x+cos^2x=1`

`LHS=(2rsinxcosx)^2+r^2(cos^2 x-sin^2 x)^2`

`=(4r^2sin^2xcos^2x)+r^2(cos^2 x-sin^2 x)^2`

`=4r^2sin^2xcos^2x+r^2(cos^4x+sin^4x-2cos^2xsin^2x)`

`=r^2(cos^4x+sin^4 x+2sin^2xcos^2x)`

`=r^2(cos^2x+sin^2x)^2`

`=r^2`

`=RHS`

`QED`

Answer 2

`LHS=(2rsinxcosx)^2 + r^2 (cos^2x -sin^2x)^2`

`=(rsin2x)^2 + r^2 (cos2x)^2`

`=r^2(sin^2 2x + cos^2 2x)`

`=r^2=RHS`

Answer 3

This question has multiple steps, so skip ahead to the explanation.

Explanation:

So, we are trying to prove: `(2rsinxcosx)^2+r^2(cos^2x-sin^2x)^2=r^2`

We can ignore the right side of the equal sign. First, let's move the `r` outside the brackets or radicals.

`r^2(2sinxcosx)^2+r^2(cos^2x-sin^2x)^2`
`r^2((2sinxcosx)^2+(cos^2x-sin^2x)^2)`

Using the proof `cos^2x-sin^2x=cos(2x)` `.^1`
`r^2((2sinxcosx)^2+(cos(2x))^2)`
`r^2((2sinxcosx)^2+cos^2(2x))`

Expand.
`r^2(4sin^2xcos^2x+cos^2(2x))`

Use the proof `cos(2x)=2cos^2x-1`
`r^2(4sin^2xcos^2x+(2cos^2x-1)^2)`
`r^2(4sin^2xcos^2x+4cos^4x-4cos^2x+1)`

Group `cosx`
`r^2(4cos^2x(sin^2x+cos^2x-1)+1)`

Use the proof `sin^2x+cos^2x=1`
`r^2(4cos^2x(1-1)+1)`
`r^2(4cos^2x(0)+1)`
`r^2(1)`
`r^2`

Comparing with the right side of the equal sign...
`r^2=r^2`.

And we're done!

Footnote 1: instead of substituting `cos^2x-sin^2x=cos(2x)` into `cos(2x)=2cos^2x-1`, you could go from `cos^2x-sin^2x` to `2cos^2x-1` without making the substitution.
`cos^2x-sin^2x`.

After rearranging the proof: `1=sin^2x+cos^2x -> -sin^2x=cos^2x-1` We can plug this in.

`cos^2x-cos^2x-1=2cos^2x-1`.