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anonymous
 3 years ago
help with verifying this please!
tanxcos^2x=2tanxcos^2xtanx/1tan^2x
anonymous
 3 years ago
help with verifying this please! tanxcos^2x=2tanxcos^2xtanx/1tan^2x

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jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0\(\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)tan(x)}{1tan^2(x)}\ \ ?\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i worked with the right side and converted everything to sin and cos first, but I just confused myself even more

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0$$\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)tan(x)}{1tan^2(x)}\\ \text{let's do the rightside first}\\ \cfrac{\frac{2sin(x)cos^2(x)}{cos(x)}\frac{sin(x)}{cos(x)}} {1\frac{sin^2(x)}{cos^2(x)}}\\ \cfrac{ \frac{2sin(x)cos^2(x)sin(x)}{cos(x)} }{ \frac{cos^2(x)sin^2(x)}{cos^2(x)}}\\ \frac{2sin(x)cos^2(x)sin(x)}{\cancel{cos(x)}} \times \frac{\cancel{cos^2(x)}}{cos^2(x)sin^2(x)} $$

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0$$\bf \frac{sin(x)(2cos^2(x)1)}{1} \times \frac{cos(x)}{cos^2(x)sin^2(x)}\\ \cfrac{sin(x)cos(x)(2cos^2(x)1)}{cos^2(x)sin^2(x)} $$

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0now take a peek at \(\bf 2cos^2(x)1 \implies cos(2x) \)

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0and \(\bf cos(2x) =cos^2(x)sin^2(x)\) which means that \(\bf 2cos^2(x)1 = cos^2(x)sin^2(x)\)

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0thus then $$\bf \cfrac{sin(x)cos(x)(2cos^2(x)1)}{cos^2(x)sin^2(x)} \implies \cfrac{sin(x)cos(x)\cancel{(cos^2(x)sin^2(x))}}{\cancel{cos^2(x)sin^2(x)}} $$

jdoe0001
 3 years ago
Best ResponseYou've already chosen the best response.0now try the left side :)
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