kryton1212
If tanΘ= -4/3 and Θ lies in the second quadrant, then sinΘ-2cosΘ=?
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gerryliyana
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sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???
kryton1212
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why we need to divided by cosΘ??
hartnn
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you can't just divide by cos theta, you need to multiply also.
hartnn
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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.
kryton1212
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i have learnt sec yet, i have learnt sin cos tan only so far
anonymous
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|dw:1365910549853:dw|
anonymous
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there is a picture of an angle whose tangent is \(\frac{4}{3}\)
anonymous
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you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle
anonymous
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|dw:1365910630356:dw|
kryton1212
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and?
anonymous
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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
anonymous
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you want
\[\sin(\theta)-2\cos(\theta)\] and you know the numbers that you need
\[\frac{4}{5}-2\times (-\frac{3}{5})=\frac{4}{5}+\frac{6}{5}\]
anonymous
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rather easy if you draw a triangle
otherwise a pita
kryton1212
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2?
hartnn
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Alternate way to look at the same thing,
|dw:1365910821672:dw|
so the angle inside triangle is \(\pi - \theta\)
so, \(\cos(\pi-\theta)=3/5, -\cos \theta=3/5 , \cos \theta = -3/5\)
anonymous
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yeah,2
kryton1212
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thank you both of you
kryton1212
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and the next questions is :
[1/cosθ + tanθ](1-sinθ)=?
hartnn
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write tan = sin/cos.
kryton1212
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i got the answer -cosθ but there are only five choices:
A. sinθ
B. cosθ
C. cos^(2)θ
D. 1+sinθ
E. sinθtanθ
hartnn
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how you got -cos theta ?
hartnn
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more specifically, how you got that - *minus*
anonymous
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\[(\frac{1}{a}+\frac{b}{a})(1-b)=\frac{1-b^2}{a}\]
Azteck
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\[\large \tan \theta=-\frac{4}{3}\]
\[\large \theta =126`52'\]
\[\sin \theta -2cos\theta=\sin 126`52' -2\cos126`52'\]
\[\large =2\]
anonymous
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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
\[\frac{\cos^2(\theta)}{\cos(\theta)}=\cos(\theta)\]
kryton1212
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|dw:1365911248263:dw|
kryton1212
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ummm, may be i got something wrong
anonymous
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the algebra is easier if you put \(\cos(\theta)=a\) and \(\sin(\theta)=b\)
anonymous
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then you won't get so confused when you multiply stuff out
almost all the steps are algebra, there is very little trig here
anonymous
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\[(\frac{1}{\cos(\theta)}+\tan(\theta))(1-\sin(\theta))\] becomes the more simple to compute
\[(\frac{1}{a}+\frac{b}{a})(1-b)\]
anonymous
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or if you prefer
\[(\frac{1+b}{a})(1-b)\]
anonymous
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now you get \(\frac{1-b^2}{a}\) almost instantly
kryton1212
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ohhh, i write sin^(2)θ-1 ....
anonymous
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actually \(1-\sin^2(\theta)\) for the numerator
kryton1212
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i want to ask (1+sinθ)(1-sinθ)=1-sin^(2)θ??
hartnn
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yes^
hartnn
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\((a+b)(a-b)=a^2-b^2\)
kryton1212
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ohhhh...then the answer is cosθ
hartnn
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yes, absolutely.
kryton1212
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and the next question:
If cosθ=1/k and 0°≤θ≤90°, then -tan(90°-θ)=?
hartnn
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|dw:1365911889934:dw|
find the unknown side by using pythagoras theorem ?
kryton1212
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k^(2)-1
hartnn
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yes.
hartnn
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|dw:1365911994361:dw|
now find (tan 90-theta) = opposite side / adjacent side
hartnn
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sorry, that should read \(\large k^2-1\)
kryton1212
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tan(90-theta)=-1/tan theta?
hartnn
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you can use that identity , its just another way to solve the same problem.
hartnn
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you'll get the same answer using any of the 2 methods.
kryton1212
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\[-\frac{ 1 }{ \sqrt{k ^{2}}-1 }\] ??
hartnn
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if you meant ,
\(\Large -\frac{ 1 }{ \sqrt{k ^{2}-1} }\)
then you are correct.
kryton1212
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thanks
hartnn
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welcome ^_^