Mathematics
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OpenStudy (superhelp101):
last question :)
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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
\[\large \tan(2\theta) = \dfrac{2\tan(\theta)}{1-\tan^2(\theta)}\]
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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
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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
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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
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ganeshie8 (ganeshie8):
Yes im getting the same!
OpenStudy (superhelp101):
yay! Thank you ganeshie!
ganeshie8 (ganeshie8):
np:)