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Identity for Cos4x
- Thread starter small1bc
- Start date
i am still really confused....how do i plug that in... we havent been given that identity yet....here is wut i have so far now...
cos(2x+2x) = cos(2x)cos(2x) - sin(2x)sin(2x)
= (4cos^4 x - 4cos^2 x +1) - (4cos^2 x - 8cos^3 x + 4cos^4 x)
but that is = 8cos^3 x - 8cos^2 x + 1, not 8cos^4 x - 8cos^2 x + 1
????????????
cos(2x+2x) = cos(2x)cos(2x) - sin(2x)sin(2x)
= (4cos^4 x - 4cos^2 x +1) - (4cos^2 x - 8cos^3 x + 4cos^4 x)
but that is = 8cos^3 x - 8cos^2 x + 1, not 8cos^4 x - 8cos^2 x + 1
????????????
Here's what I was getting at. It's a little overdone compared to Soroban, but maybe you can see the use of identities.
\(\displaystyle \L\\8cos^{4}(x)-8cos^{2}+1\)
\(\displaystyle \L\\8cos^{2}(x)(cos^{2}(x)-1)+1\)
\(\displaystyle \L\\8cos^{2}(x)(-sin^{2}(x))+1\)
\(\displaystyle \L\\8\left(\frac{1}{2}+\frac{cos(2x)}{2}\right)\left(\frac{cos(2x)}{2}-\frac{1}{2}\right)+1\)
\(\displaystyle \L\\8\left(\frac{cos^{2}(2x)-1}{4}\right)+1\)
\(\displaystyle \L\\8\left(\frac{1+cos(4x)}{8}-\frac{1}{4}\right)+1\)
\(\displaystyle \L\\(1+cos(4x)-2)+1\)
\(\displaystyle \H\\cos(4x)\)
\(\displaystyle \L\\8cos^{4}(x)-8cos^{2}+1\)
\(\displaystyle \L\\8cos^{2}(x)(cos^{2}(x)-1)+1\)
\(\displaystyle \L\\8cos^{2}(x)(-sin^{2}(x))+1\)
\(\displaystyle \L\\8\left(\frac{1}{2}+\frac{cos(2x)}{2}\right)\left(\frac{cos(2x)}{2}-\frac{1}{2}\right)+1\)
\(\displaystyle \L\\8\left(\frac{cos^{2}(2x)-1}{4}\right)+1\)
\(\displaystyle \L\\8\left(\frac{1+cos(4x)}{8}-\frac{1}{4}\right)+1\)
\(\displaystyle \L\\(1+cos(4x)-2)+1\)
\(\displaystyle \H\\cos(4x)\)