Question

Question #8e0f7

Answer 1

See the Proof in Explanation.

Explanation:

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

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

`cos(x+x)=cosx*cosx-sinx*sinx`

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

Hence, the Proof.

Is it helpful? Enjoy Maths.!

Answer 2

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`