Precalculus
7 Online
OpenStudy (anonymous):
last one prove the identity show all work 2tanx /1+tan^2x=sin2x
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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
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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):
\(\huge 1+ \frac{sin^2x}{cos^2x }= \frac{sin^2x+cos^2x}{cos^2x}=?\)
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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 = ?
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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
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hartnn (hartnn):
welcome ^_^
sweet dreams :)
OpenStudy (anonymous):
thanks same to you