Question #8e0f7

2 Answers
Jul 5, 2016

See the Proof in Explanation.

Explanation:

We use the Formula : cos(A+B)=cosAcosB-sinASinB.:cos(A+B)=cosAcosBsinAsinB.

Letting A=B=xA=B=x, we get,

cos(x+x)=cosx*cosx-sinx*sinxcos(x+x)=cosxcosxsinxsinx

:. cos2x=cos^2x-sin^2x, or, sin^2x+cos2x=cos^2x.

Hence, the Proof.

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Jul 5, 2016

See below.

Explanation:

Answering this question requires the use of two important identities:

  • sin^2x+cos^2x=1-> Pythagorean Identity
  • cos2x=cos^2x-sin^2x-> Double angle identity for cosine

Note that subtracting cos^2x from both sides in the first identity yields sin^2x=1-cos^2x, and it's this modified form of the Pythagorean Identity we will be using.

Now that we have a few identities to work with, we can do some substituting in sin^2x+cos2x=cos^2x:
underbrace(1-cos^2x)+underbrace(cos^2x-sin^2x)=cos^2x
color(white)Xsin^2xcolor(white)(XXXXX)cos2x

We see that the cosines cancel:
1-cancel(cos^2x)+cancel(cos^2x)-sin^2x=cos^2x
->1-sin^2x=cos^2x

This is another form of the Pythagorean Identity sin^2x+cos^2x=1; see what happens you you subtract sin^2x from both sides:
sin^2x+cos^2x=1
sin^2x+cos^2x-sin^2x=1-sin^2x
cancel(sin^2x)+cos^2x-cancel(sin^2x)=1-sin^2x
->cos^2x=1-sin^2x

That's exactly what we have in 1-sin^2x=cos^2x, so we can complete the proof:
cos^2x=cos^2x