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Precalculus 19 Online
OpenStudy (anonymous):

last one prove the identity show all work 2tanx /1+tan^2x=sin2x

hartnn (hartnn):

thats 1+\(tan^2x\) in denominator, right ?

hartnn (hartnn):

else it can't be proved

OpenStudy (anonymous):

yes sorry it was my mistake

hartnn (hartnn):

so write tanx in terms of sin x and cos x, both in num and denom

OpenStudy (anonymous):

so 2sinx/cosx/1+cos^2x/sin^2x

hartnn (hartnn):

yeah, simplify it

OpenStudy (anonymous):

idk how to do that

hartnn (hartnn):

lets take denominator only \(1+ \frac{sin^2x}{cos^2x }= ? \)

OpenStudy (anonymous):

okay

hartnn (hartnn):

\(\huge 1+ \frac{sin^2x}{cos^2x }= \frac{sin^2x+cos^2x}{cos^2x}=?\)

OpenStudy (anonymous):

okay

hartnn (hartnn):

what numerator equals there ^ ?

OpenStudy (anonymous):

1

hartnn (hartnn):

yes, so in denominator , its only 1/cos^2 x now what is 2sin x/ cos x /(1/cos^2 x ) = ?

hartnn (hartnn):

divide by 1/cos^2 x = multiply by cos^2 x so 2sin x/ cos x /(1/cos^2 x ) = 2 sin x cos^2x / cos x = ?

OpenStudy (anonymous):

2sinxcos^2 x?

hartnn (hartnn):

one cos x will cancel from numerator, right ?

OpenStudy (anonymous):

yea

hartnn (hartnn):

and sin 2x = 2sin x cos x is a standard formula

OpenStudy (anonymous):

ok thank you so much thats all for today thanks for ur help have a great night

hartnn (hartnn):

welcome ^_^ sweet dreams :)

OpenStudy (anonymous):

thanks same to you

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