vera_ewing
  • vera_ewing
Math question
Mathematics
chestercat
  • chestercat
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vera_ewing
  • vera_ewing
Michele_Laino
  • Michele_Laino
hint: \[\sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} \]
Michele_Laino
  • Michele_Laino
hint: \[\large \begin{gathered} \sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} = \pm \sqrt {1 - \frac{{16}}{{49}}} = \pm \sqrt {\frac{{33}}{{49}}} = ...? \hfill \\ \hfill \\ \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \sqrt {\frac{{33}}{{49}}} } \right)}}{{ - \frac{4}{7}}} = ...? \hfill \\ \end{gathered} \]

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vera_ewing
  • vera_ewing
I'm not sure...
Michele_Laino
  • Michele_Laino
hint: \[ \pm \sqrt {\frac{{33}}{{49}}} = \pm \frac{{\sqrt {33} }}{7}\]
vera_ewing
  • vera_ewing
@Michele_Laino Okay, so sin0 = (sqrt)33/7 ?
anonymous
  • anonymous
not to butt in but this is somewhat easier if you draw a triangle
anonymous
  • anonymous
|dw:1436190352602:dw|
anonymous
  • anonymous
there is picture of an angle whose cosine is \(\frac{4}{7}\) find the "opposite" side via pythagoras then you can take any trig ratio you like
Michele_Laino
  • Michele_Laino
since we have 2 square roots, then: \[\sin \theta = \pm \frac{{\sqrt {33} }}{7}\]
vera_ewing
  • vera_ewing
How do we find tanθ Michele?
Michele_Laino
  • Michele_Laino
it is simple, you have to compute this ratio: \[\large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \frac{{\sqrt {33} }}{7}} \right)}}{{ - \frac{4}{7}}} = \left( { \pm \frac{{\sqrt {33} }}{7}} \right) \times \left( { - \frac{7}{4}} \right) = ...?\]
vera_ewing
  • vera_ewing
-7(sqrt)33/28 @Michele_Laino ?
anonymous
  • anonymous
|dw:1436190696248:dw|
anonymous
  • anonymous
memorize the trig ratios, look at the triangle and you can find any trig ratio you like
anonymous
  • anonymous
for example \[\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\pm\frac{\sqrt{33}}{7}\]
anonymous
  • anonymous
\[\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=?\]
anonymous
  • anonymous
there is really very very little work to solve these just find the missing side of the triangle and use the trig ratios you already know
vera_ewing
  • vera_ewing
tanθ = (sqrt)33/4 right?
anonymous
  • anonymous
yes of course
vera_ewing
  • vera_ewing
Thank you.
anonymous
  • anonymous
yw

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