anonymous
  • anonymous
sin theta= 3/5 cos beta = 2sqrt6/5 Find sin(theta+beta). Is the answer 6sqrt6+4/25?
Mathematics
chestercat
  • chestercat
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shamim
  • shamim
R u able to find out Cos theta=?
anonymous
  • anonymous
Yes 4/5
UsukiDoll
  • UsukiDoll
\[\sin (\theta) = \frac{3}{5}, \cos(\beta) = 2\sqrt{\frac{6}{5}}\] ? I'm confused on what cosine beta should be

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shamim
  • shamim
Sin beta=?
anonymous
  • anonymous
1/5
UsukiDoll
  • UsukiDoll
oh yeah we need sin beta and cosine beta otherwise the angle sum identity formula isn't going to work
UsukiDoll
  • UsukiDoll
\[\sin(\theta + \beta) = \sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta)\]
UsukiDoll
  • UsukiDoll
so far we have sin (theta) = 3/5 |dw:1433043667731:dw| so cos (theta) = 4/5
anonymous
  • anonymous
Yep got that
UsukiDoll
  • UsukiDoll
ummm what was the cosine beta?
UsukiDoll
  • UsukiDoll
is it \[\cos(\beta) = 2 \frac{\sqrt{6}}{5} ?\] is the square root inside the 6/5? that part is throwing me off
UsukiDoll
  • UsukiDoll
\[\sin(\theta + \beta) = \frac{3}{5}\cos(\beta)+\frac{4}{5}\sin(\beta)\] I don't understand what cosine beta is... but I do know that I need another triangle so I can grab sin beta and we add this all together
UsukiDoll
  • UsukiDoll
is the 6/5 fraction inside the whole square root or is it just square root 6 / 5 ?
UsukiDoll
  • UsukiDoll
\[\cos(\beta) = 2 \frac{\sqrt{6}}{5} ? \] or \[\cos(\beta) = 2 \sqrt{\frac{6}{5}} \]
UsukiDoll
  • UsukiDoll
I need clarification on that part... otherwise I might end up with the wrong answer.
UsukiDoll
  • UsukiDoll
|dw:1433044545197:dw|
UsukiDoll
  • UsukiDoll
\[\sin(\theta + \beta) = \frac{3}{5}\frac{2\sqrt{6}}{5}+\frac{4}{5} (\frac{1}{5})\]
UsukiDoll
  • UsukiDoll
\[\frac{6\sqrt{6}}{25}+\frac{4}{25}\]
UsukiDoll
  • UsukiDoll
I relooked at cos beta and realized that 5 was the longest side of the triangle and that 2 sqrt{6} was the adjcent part... needed to find the opposite part of the triangle so I could find sin beta.
UsukiDoll
  • UsukiDoll
you got the right answer.

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