Question #12b6c

3 Answers
Sep 22, 2017

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

Sep 22, 2017

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

Sep 22, 2017

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.