Question

Formula for (1-cos x)? ​

Answer 1

Answer:

There isn’t one formula, there are several.

1−cos(x)=[1−cos(x)].1+cos(x)1+cos(x)1−cos⁡(x)=[1−cos(x)].1+cos⁡(x)1+cos⁡(x)

=1−cos2(x)1+cos(x)=sin2(x)1+cos(x)=1−cos2⁡(x)1+cos⁡(x)=sin2⁡(x)1+cos⁡(x)

Ok, perhaps this isn’t that useful, but from the half-angle formula for the cosinecosine function, we can rewrite this as:

sin2(x)2cos2(0.5x)sin2⁡(x)2cos2⁡(0.5x)

While we are about it, let’s use the half-angle formula for the sinesine function:

4sin2(0.5x)cos2(0.5x)2cos2(0.5x)=2sin2(0.5x)4sin2⁡(0.5x)cos2⁡(0.5x)2cos2⁡(0.5x)=2sin2⁡(0.5x)

We could express cos(x)cos⁡(x) in terms of t=tan(0.5x)t=tan⁡(0.5x)

1−cos(x)=1−1−t21+t2=2t21+t21−cos⁡(x)=1−1−t21+t2=2t21+t2

=tan(0.5x)sin(x)=2sin2(0.5x)=tan⁡(0.5x)sin⁡(x)=2sin2⁡(0.5x)

How about something different?

cos(x)=∑∞n=0(−1)nx2n(2n)!cos⁡(x)=∑n=0∞(−1)nx2n(2n)!

Thus

1−cos(x)=−∑∞n=1(−1)nx2n(2n)!1−cos⁡(x)=−∑n=1∞(−1)nx2n(2n)!

=x22!−x44!+x66!...

Answer 2

Answer:

Depend on question. 1- 1/sin x