#Double-Angle Identity
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I have found the derivatives using the product rule but I don't understand how I can use the identities to find the derivatives
do you know addition laws in trig?
I know the cos and sin laws but not tan
should be cos^2x
ah
i dont see how
that would imply 1 - sin^2x = 1 - 2sin^2x for all real x which isnt true
I came across this explanation but it doesn't make sense to me
We have,
cos2x = cos^2x - sin^2x = cos^2x - (1 - cos^2x) [Because cos^2x + sin^2x = 1 ⇒ sin^2x = 1 - cos^2x] = cos^2x - 1 + cos^2x = 2cos^2x - 1
Hence , we have cos2x = 2cos^2x - 1 in terms of cosx
sin2x is not 1 -cos2x
Oh the squared power looks like a regular 2, let me edit it
the first identity is wrong it's prolly a typo, it should say cos2x = 1-2sin^2(x)
but here's the derivation
cos(2x)
= cos(x+x)
= cos(x) • cos(x) - sin(x) • sin(x)
= cos^2(x) - sin^2(x)
= 1 - sin^2(x) - sin^2(x)
= 1 - 2sin^2(x)
(we can also rewrite this as 2cos^2(x) - 1 which will be useful for the next derivation)
similar process:
cos(3x)
= cos(2x+x)
= cos(2x) • cos(x) - sin(2x) • sin(x)
= [2cos^2(x) - 1] • [cos(x)] -
[2 • sin(x) • cos(x)] • [sin(x)]
= 2 • cos^3(x) - cos(x) - 2 • sin^2(x) • cos(x)
= 2 • cos^3(x) - cos(x) -
2 • [1-cos^2(x)] • cos(x)
= 2 • cos^3(x) - cos(x) - 2 • cos(x) +
2 • cos^3(x)
= 4cos^3(x) - 3cos(x)
finding the derivative of sin^2(x) by
- product rule:
sin^2(x) = sin(x) • sin(x)
by the product rule, we have:
cos(x) • sin(x) + sin(x) • cos(x)
= 2 • sin(x) • cos(x)
= sin(2x)
- using identities
sin^2(x)
= 1/2 • [1-cos(2x)]
differentiating this, we have:
1/2 • [0 + 2 • sin(2x)]
= 1/2 • [2 • sin(2x)]
= sin(2x)
okay that took ages to type on my phone u get the point, it'll be the same thing for differentiating cos^3(x)
Thank you so much. It makes so much sense
im glad, no worries!!
no
cos2x = 1-2sin^2x
GET THIS GUY OFF HIS MATHS DEGREE