Help verifying Trig Identity: Sin2x = 2CotXSin^2x
- Thread starter Jaskaran
- Start date
\(\displaystyle \L\ sin(nx) = n!sin(x)\sum_{k=0}^n \frac{(-1)^k cos^{n - 2k - 1}(x) sin^{2k}(x)}{(n - 2k - 1)!(2k + 1)!}\\\)
If we put n = 2, then:
\(\displaystyle \L\ sin(2x) = 2sin(x)cos(x)\)
\(\displaystyle \L\ 2cot(x)sin^2(x) = 2(\frac{cos(x)}{sin(x)}\)sin^2(x) = 2sin(x)cos(x)\), \(\displaystyle \L\ x \not=\ k\pi\\), \(\displaystyle \L\ k \in\ J\)
\(\displaystyle \L\ 1 - cos^2(x) = sin^2(x)\)
If we put n = 2, then:
\(\displaystyle \L\ sin(2x) = 2sin(x)cos(x)\)
\(\displaystyle \L\ 2cot(x)sin^2(x) = 2(\frac{cos(x)}{sin(x)}\)sin^2(x) = 2sin(x)cos(x)\), \(\displaystyle \L\ x \not=\ k\pi\\), \(\displaystyle \L\ k \in\ J\)
\(\displaystyle \L\ 1 - cos^2(x) = sin^2(x)\)
Work on the right hand side:
\(\displaystyle \L 2Cot(x) = 2\frac{1}{tan} \cdot sin^{2}(x)\)
\(\displaystyle \L 2\frac{1}{tan(x)} = \frac{2}{\frac{sin(x)}{cos(x)}} = 2 \frac{cos(x)}{sin(x)}\) (multiplying by the reciprocal to get ride of the division)
so now the entire right side we have:
\(\displaystyle \L [2\frac{cos(x)}{sin(x)}\, \cdot \, \frac{Sin(x)Sin(x)}{1}] = \frac{2cos(x)sin(x)sin(x)}{sin(x)} = 2cos(x)sin(x)\) (we canceled out one sin(x) from the top and bottom)
Now you know your double angle trig formula \(\displaystyle \L Sin(2x) = 2sin(x)cos(x)\)
And you're proof is done!
See?
John
