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kryton1212
Group Title
If tanΘ= 4/3 and Θ lies in the second quadrant, then sinΘ2cosΘ=?
 one year ago
 one year ago
kryton1212 Group Title
If tanΘ= 4/3 and Θ lies in the second quadrant, then sinΘ2cosΘ=?
 one year ago
 one year ago

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gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
sinΘ2cosΘ = (sin Θ/ cos Θ) (2cos Θ/cosΘ) = tan Θ  2 = 4/3  2 = ???
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
why we need to divided by cosΘ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you can't just divide by cos theta, you need to multiply also.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
i would suggest you find sec theta first, from the identity sec^2 theta = 1+ tan^2 theta and then, cos = 1/ sec, sin = tan * cos.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i have learnt sec yet, i have learnt sin cos tan only so far
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
dw:1365910549853:dw
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
there is a picture of an angle whose tangent is \(\frac{4}{3}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you need the third side, which you get via pythagoras or by remembering the 3  4  5 right triangle
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
dw:1365910630356:dw
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now you see that \(\sin(\theta)=\frac{4}{5}\) and \(\cos(\theta)=\frac{3}{5}\) except that since you are in quadrant II you have \(\cos(\theta)=\frac{3}{5}\) because in quadrant II cosine is negative
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you want \[\sin(\theta)2\cos(\theta)\] and you know the numbers that you need \[\frac{4}{5}2\times (\frac{3}{5})=\frac{4}{5}+\frac{6}{5}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
rather easy if you draw a triangle otherwise a pita
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
Alternate way to look at the same thing, dw:1365910821672:dw so the angle inside triangle is \(\pi  \theta\) so, \(\cos(\pi\theta)=3/5, \cos \theta=3/5 , \cos \theta = 3/5\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yeah,2
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
thank you both of you
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
and the next questions is : [1/cosθ + tanθ](1sinθ)=?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
write tan = sin/cos.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i got the answer cosθ but there are only five choices: A. sinθ B. cosθ C. cos^(2)θ D. 1+sinθ E. sinθtanθ
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
how you got cos theta ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
more specifically, how you got that  *minus*
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[(\frac{1}{a}+\frac{b}{a})(1b)=\frac{1b^2}{a}\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
\[\large \tan \theta=\frac{4}{3}\] \[\large \theta =126`52'\] \[\sin \theta 2cos\theta=\sin 126`52' 2\cos126`52'\] \[\large =2\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
in this case \(a=\cos(\theta)\) and \(b=\sin(\theta)\) so you get \[\frac{1\sin^2(\theta)}{\cos(\theta)}\] since \(1\sin^2(\theta)=\cos^2(\theta)\) you end up with \[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
dw:1365911248263:dw
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ummm, may be i got something wrong
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
then you won't get so confused when you multiply stuff out almost all the steps are algebra, there is very little trig here
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[(\frac{1}{\cos(\theta)}+\tan(\theta))(1\sin(\theta))\] becomes the more simple to compute \[(\frac{1}{a}+\frac{b}{a})(1b)\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
or if you prefer \[(\frac{1+b}{a})(1b)\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now you get \(\frac{1b^2}{a}\) almost instantly
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ohhh, i write sin^(2)θ1 ....
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
actually \(1\sin^2(\theta)\) for the numerator
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
i want to ask (1+sinθ)(1sinθ)=1sin^(2)θ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\((a+b)(ab)=a^2b^2\)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
ohhhh...then the answer is cosθ
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yes, absolutely.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
and the next question: If cosθ=1/k and 0°≤θ≤90°, then tan(90°θ)=?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
dw:1365911889934:dw find the unknown side by using pythagoras theorem ?
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
k^(2)1
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
dw:1365911994361:dw now find (tan 90theta) = opposite side / adjacent side
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
sorry, that should read \(\large k^21\)
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
tan(90theta)=1/tan theta?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you can use that identity , its just another way to solve the same problem.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you'll get the same answer using any of the 2 methods.
 one year ago

kryton1212 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ \sqrt{k ^{2}}1 }\] ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
if you meant , \(\Large \frac{ 1 }{ \sqrt{k ^{2}1} }\) then you are correct.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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