Wednesday, October 25, 2006

cos(z) = sin(π/2 + z) and other identities

In today's blog, I go through a through a set of addition properties of sin, cos, tan, and cot.

Lemma 1: cos(z) = sin(z + π/2)

Proof:

(1) sin(π/2 + z) = sin(π/2)cos(z) + sin(z)cos(π/2)

(2) sin(π/2) = 1, cos(π/2)=0 [See Property 2, Property 7, here, see here for explanation on radians if needed]

(3) This gives us:

sin(π/2 + z) = cos(z)*1 + sin(z)*0.

QED

Corollary 1.1: cos(π/2 + x) = -sin(x)

Proof:

(1) -sin(x) = -sin(x + π/2 - π/2) = sin(-x + π/2 - π/2) [See Property 4, here]

(2) sin(π/2 + [-x - π/2]) = cos(-x -π/2) [From Lemma 1 above]

(3) cos(-x -π/2) = cos(x + π/2) [See Property 9, here]

QED

Corollary 1.2: sin(x + π) = -sin(x)

Proof:

(1) sin(x + π) = sin(x + π/2 + π/2) = cos(x + π/2) [See Lemma 1 above]

(2) cos(x + π/2) = -sin(x) [See Corollary 1.1 above]

QED

Corollary 1.3: cos(x + π) = -cos(x)

Proof:

(1) cos(x + π) = cos(x + π/2 + π/2) = -sin(x + π/2) [See Corollary 1.1 above]

(2) -sin(x + π/2) = -cos(x) [See Lemma 1 above]

QED

Corollary 1.4: tan(x + π) = tan(x)

Proof:

(1) tan(x + π) = sin(x + π)/cos(x + π) = -sin(x)/[-cos(x)] = sin(x)/cos(x) = tan(x) [See Corollary 1.2 and 1.3 above]

QED

Corollary 1.5: cot(x + π) = cot(x)

Proof:

(1) cot(x + π) = 1/tan(x + π) = 1/tan(x) = cot(x) [See Corollary 1.4 above]

QED

Corollary 1.6: tan(x + π/2) = -cot(x)

Proof:

(1) tan(x + π/2) = sin(x + π/2)/cos(x + π/2) = cos(x)/[-sin(x)] = -cot(x) [See Lemma 1 and Corollary 1.1 above]

QED

Corollary 1.7: sin(x) = cos(π/2 - x)

Proof:

(1) cos(π/2 - x) = cos(π/2)cos(-x) - sin(π/2)sin(-x) [See Theorem 2, here]

(2) cos(-x) = x [See Property 9, here]

(3) sin(-x) = -sin(x) [See Property 4, here]

(4) cos(π/2)cos(-x) - sin(π/2)sin(-x) = cos(90°)cos(x) + sin(90°)sin(x)

(5) cos(90°) = 0 [See Property 7, here]

(6) sin(90°) = 1 [See Property 2, here]

(7) cos(π/2 - x) = 0*cos(x) + 1*sin(x) = sin(x)

QED

Corollary 1.8: cos(x) = sin(π/2 - x)

Proof:

(1) sin(x) = cos(π/2 - x) [See Corollary 1.7 above]

(2) Let x = π/2 - z

(3) sin(π/2 - z) = cos(π/2 - [π/2 - z]) = cos(z)

QED

Lemma 2: sin(2x) = 2sin(x)cos(x)

Proof:

(1) sin(x + x) = sin(x)cos(x) + cos(x)sin(x) [See Theorem 1, here]

(2) sin(2x) = 2sin(x)cos(x)

QED

Lemma 3: cos(2x) = cos2(x) - sin2(x)

Proof:

(1) cos(x + x) = cos(x)*cos(x) - sin(x)*sin(x) [See Theorem 2, here]

(2) cos(2x) = cos2(x) - sin2(x)

QED

Corollary 3.1: cos(2x) = 2cos2(x) - 1 = 1 - 2sin2(x)

Proof:

(1) cos(2x) = cos2(x) - sin2(x) [See Lemma 3 above]

(2) cos2(x) + sin2(x) = 1. [See Corollary 2, here]

(3) From step #2, we have:

cos2(x) - 1 = -sin2(x)

and also

cos2(x) = 1 - sin2(x)

(3) Combining step #3 and step #1 gives us:

cos(2x) = (1 - sin2(x)) - sin2(x) = 1 - 2sin2(x)

cos(2x) = cos2(x) + (cos2 - 1) = 2cos2(x) - 1

QED

Corollary 3.2: sin2(x) = [1 - cos(2x)]/2

Proof:

(1) cos(2x) = 1 - 2sin2(x) [See Corollary 3.1 above]

(2) 2sin2(x) = 1 - cos(2x)

(3) So that we have:

sin2(x) = [1 - cos(2x)]/2

QED

Corollary 3.3: cos2(x) = [1 + cos(2x)]/2

Proof:

(1) cos(2x) = 2cos2(x) - 1 [See Corollary 3.1 above]

(2) 2cos2(x) = cos(2x) + 1

(3) cos2(x) = [cos(2x) + 1]/2

QED

Corollary 3.4: cos(x/2) = ± √(1/2)[1 + cos(x)]

Proof:

(1) cos2(x/2) = [1 + cos(x)]/2. [From Corollary 3.3 above]

(2) cos(x/2) = ± √(1/2)[1 + cos(x)]

QED

Corollary 3.5: sin(x/2) = ± √(1/2)[1 - cos(x)]

Proof:

(1) sin2(x/2) = [1 - cos(x)]/2. [From Corollary 3.2 above]

(2) sin(x/2) = ± √(1/2)[1 - cos(x)]

QED

Corollary 3.6: cot(2x) = (1/2)[cot(x) - tan(x)]

Proof:

(1) cot(2x) = cos(2x)/sin(2x) = [cos2(x) - sin2(x)]/(2sin(x)cos(x)) =

= (1/2)[cos(x)/sin(x) - sin(x)/cos(x)] = (1/2)[cot(x) - tan(x)]

QED

References

2 comments :

Joe Whitehead said...

"Corollary 3.2: sin2(x) = [1 - cos(2x)]/2
Proof:
(1) cos(2x) = 1 - 2sin2(x) [See Corollary 3.1 above]
(2) 2sin2(x) = 1 - cos(2x)
(3) So that we have:
sin2(x) = [1 + cos(2x)]/2
QED"

Is that a typo? Shouldn't it be the same as above in #3?

Larry Freeman said...

Hi Joe,

Yes, that's a typo. I just fixed it.

Thanks very much for noticing that.

Cheers,

-Larry