Proving the Power-Reducing and Half-Angle Trig Identities

In this discussion, we are going to prove the Power-Reducing and Half-Angle Trig Identities.

Here is an outline of this discussion:

🔹Proof of the Power-Reducing Identities

Proving sin^2x = [1- cos2x]/2

Proving cos^2x = [cos2x+ 1 ] / 2

Proving tan^2x = [1 – cos2x] / [1+ cos2x]

🔹Proof of the Half-Angle Identities

Proving sin(x/2) = +- sqrt([1- cosx]/2)

Proving cos(x/2) = +- sqrt([cosx+1]/2)

Proving tan(x/2) = +- sqrt([1 – cosx] / [1+ cosx])

Let’s begin!

Proof of the Power-Reducing Identities

Proving sin^2x = [1- cos2x]/2

We begin using the double-angle identities that we proof in previous discussions. We know by the double-angle identity that:

cos2x = 1 – 2sin^2x

Solving for sin^2x, we have

cos2x = 1- 2sin^2x

cos2x -1 = – 2sin^2x

1 – cos2x = 2sin^2x

[1- cos2x ]/2 = sin^2x

Thus,

sin^2x = [1- cos2x ]/2

Proving cos^2x = [cos2x + 1 ] / 2

Similarly, we know by the double-angle identity that:

cos2x = 2cos^2x – 1

Solving for cos^2x, we have:

cos2x = 2cos^2x – 1

cos2x + 1 = 2cos^2x

[cos2x + 1 ] / 2 = cos^2x

Thus,

cos^2x = [cos2x + 1 ] / 2

Proving tan^2x = [1 – cos2x ] / [cos2x + 1]

We have proved that sin^2x = [1- cos2x]/2 and cos^2x = [cos2x + 1 ] / 2.

So,

tan^2x = sin^2x / cos^2x

= [1- cos2x]/2 / [cos2x + 1 ] / 2

= [1- cos2x]/2 * 2/ [cos2x + 1 ]

= [1- cos2x] / [cos2x + 1 ]

Thus,

tan^2x = [1 – cos2x] / [1+cos2x]

Proof of the Half-Angle Identities

Proving sin(x/2) = +- sqrt([1- cosx]/2)

We have shown that:

sin^2x = [1- cos2x]/2

Taking the square root of both sides, we have

sinx = +- sqrt([1- cos2x]/2)

Thus,

sin(x/2) = +- sqrt([1- cos(2x/2)]/2)

sin(x/2) = +- sqrt([1- cosx]/2)

The positive and negative root depend upon the quadrant in which x/2 lies.

Proving cos(x/2) = +- sqrt([cosx+1]/2)

We have shown that

cos^2x = [cos2x+ 1 ] / 2

Taking the square root of both sides, we have

cos^2x = [cos2x+ 1 ] / 2

cosx = +-sqrt([cos2x+ 1 ] / 2)

thus,

cos(x/2) = +-sqrt([cos(2x/2) + 1 ] / 2)

cos(x/2) = +-sqrt([cosx + 1 ] / 2)

The positive and negative root depend upon the quadrant in which x/2 lies.

Proving tan(x/2) = +- sqrt([1-cosx]/[1+cosx])

We have shown that

tan^2x = [1 – cos2x] / [1+ cos2x]

Taking the square root of both sides, we have

tan^2x = [1 – cos2x] / [1 + cos2x ]

tanx = +- sqrt([1 – cos2x] / [1 + cos2x ])

thus,

tan(x/2) = +- sqrt([1 – cos(2x/2)] / [1 + cos(2x/2)])

tan(x/2) = +- sqrt([1 – cos(x)] / [1 + cos(x)])

The positive and negative root depend upon the quadrant in which x/2 lies.

We are done!

Thank you for reading!

I hope that I can assist you on your math journey!

Sincerely,

➕MathAngel369➕

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