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Mathematics 21 Online
OpenStudy (superhelp101):

last question :)

OpenStudy (superhelp101):

What is tan2(theta), if theta is a second quadrant angle with sin(theta)=4/5

OpenStudy (superhelp101):

@ganeshie8

ganeshie8 (ganeshie8):

you mean \(\large \tan(2\theta)\) or \(\large \tan^2(\theta)\) ?

OpenStudy (superhelp101):

first one

ganeshie8 (ganeshie8):

good, use double angle formula \[\large \tan(2\theta) = \dfrac{2\tan(\theta)}{1-\tan^2(\theta)}\]

ganeshie8 (ganeshie8):

find the value of \(\tan(\theta)\) and plug it in abvoe

OpenStudy (superhelp101):

do you think you can show me ? :)

ganeshie8 (ganeshie8):

draw a triangle based on given info : \(\large \sin(\theta) = \dfrac{4}{5}\)

ganeshie8 (ganeshie8):

|dw:1430197274494:dw|

OpenStudy (superhelp101):

-3 is the other side

ganeshie8 (ganeshie8):

Yes! so \(\tan(\theta) = ?\)

OpenStudy (superhelp101):

53.13

ganeshie8 (ganeshie8):

\(\tan(\theta)\) is a ratio of opposite side and adjacent side not angle..

ganeshie8 (ganeshie8):

try again

OpenStudy (superhelp101):

would be 4/-3

ganeshie8 (ganeshie8):

Yes!

ganeshie8 (ganeshie8):

\[\tan(\theta) = -\dfrac{4}{3}\] simply plug that in the earlier identity of \(\tan(2\theta)\)

ganeshie8 (ganeshie8):

\[\large\begin{align} \tan(2\theta) &= \dfrac{2\tan(\theta)}{1-\tan^2(\theta)}\\~\\&=\dfrac{2\left(-\frac{4}{3}\right)}{1-\left(-\frac{4}{3}\right)^2}\\~\\&=?\end{align}\]

OpenStudy (superhelp101):

wait I think i got it; it it 24/7

OpenStudy (superhelp101):

it should be 24/7

ganeshie8 (ganeshie8):

Yes im getting the same!

OpenStudy (superhelp101):

yay! Thank you ganeshie!

ganeshie8 (ganeshie8):

np:)

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