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

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

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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1i 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.

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

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

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1now 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

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

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

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

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

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

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

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

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1in 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 \[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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