anonymous
  • anonymous
Need help with some trig problems
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Use the identity cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)

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jim_thompson5910
  • jim_thompson5910
you're given tan(B). Use that to find the value of sin(B)
anonymous
  • anonymous
wait we are solving question number 1 right
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
ok how do I do that
jim_thompson5910
  • jim_thompson5910
draw a right triangle with the angle beta |dw:1438649540480:dw|
jim_thompson5910
  • jim_thompson5910
if tan(B) = -1/2, then what does that mean for the right triangle above?
anonymous
  • anonymous
ok one sec
anonymous
  • anonymous
sin is \[\sqrt{5}/5\]
jim_thompson5910
  • jim_thompson5910
correct |dw:1438649735884:dw|
jim_thompson5910
  • jim_thompson5910
what's cos(B) equal to?
anonymous
  • anonymous
\[\frac{ 2\sqrt{5} }{ 5 }\]
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
actually, B is in Q2, so cosine is negative
jim_thompson5910
  • jim_thompson5910
now you do the same for the angle alpha |dw:1438649901995:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438649921905:dw|
anonymous
  • anonymous
oh yeah
jim_thompson5910
  • jim_thompson5910
sin(alpha) will be negative since alpha is in Q4
anonymous
  • anonymous
\[\sqrt{23}/5\]
jim_thompson5910
  • jim_thompson5910
that will be negative, but otherwise, good
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
\[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) =??\]
anonymous
  • anonymous
so \[\frac{ \sqrt{2} }{ 5 }\times \frac{ -2\sqrt{5} }{ 5 }+-\frac{ \sqrt{23} }{ 5 }\times- \frac{\sqrt{5} }{ 5 }\]
jim_thompson5910
  • jim_thompson5910
sin(B) is positive because B is in Q2
anonymous
  • anonymous
I cant get an exact answer on my calculator
jim_thompson5910
  • jim_thompson5910
\[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) = \frac{-2\sqrt{2*5}}{5*5} + \frac{-\sqrt{23*5}}{5*5}\] \[\Large \cos(\alpha - \beta) = -\frac{2\sqrt{10}}{25} - \frac{\sqrt{115}}{25}\]
jim_thompson5910
  • jim_thompson5910
Optionally, you can factor out the GCF \(\Large \frac{\sqrt{5}}{25}\) to get \[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) = \frac{\sqrt{5}}{25}\left(-2\sqrt{2}-\sqrt{23}\right)\]
anonymous
  • anonymous
\[\frac{ -2\sqrt{10}-\sqrt{115} }{ 25 }\]
anonymous
  • anonymous
is that the answer
jim_thompson5910
  • jim_thompson5910
yeah that's another way to write the answer
anonymous
  • anonymous
ok how do we do the second one
jim_thompson5910
  • jim_thompson5910
review this list of identities http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look for the "tan(2theta)" identity on the second page
anonymous
  • anonymous
ok i see it
jim_thompson5910
  • jim_thompson5910
so you need to find tan(alpha) and plug it into that identity
anonymous
  • anonymous
ok
anonymous
  • anonymous
tan is \[\frac{ - \sqrt{46}}{ 2}\]
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
alpha is in Q4 where tan is negative
anonymous
  • anonymous
ok so can you show me how you plug it in
jim_thompson5910
  • jim_thompson5910
\[\Large \tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}\] \[\Large \tan(2\alpha) = \frac{2\left(\frac{-\sqrt{46}}{2}\right)}{1-\left(\frac{-\sqrt{46}}{2}\right)^2}\] \[\Large \tan(2\alpha) = \frac{-\sqrt{46}}{1-\frac{46}{4}}\] I'll let you simplify
anonymous
  • anonymous
ok i got\[\frac{ 2\sqrt{46} }{ 21 }\]
jim_thompson5910
  • jim_thompson5910
me too
anonymous
  • anonymous
great does the attachement you sent me have the equation for the third one as well
jim_thompson5910
  • jim_thompson5910
yes, look under "Sum and Difference Formulas" on page 2
anonymous
  • anonymous
sin(a)cos(b)+cos(a)sin(b)
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
ok again I am having problems plugging the equations in here from my side, can you show me what you got when you plugged each of the values in
jim_thompson5910
  • jim_thompson5910
The values of sin(alpha), sin(beta), cos(alpha), cos(beta) were determined previously. Plug them in to get \[\Large \sin(\beta + \alpha) = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)\] \[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) =??\]
anonymous
  • anonymous
I am having a hard time typing it in
anonymous
  • anonymous
one second
anonymous
  • anonymous
\[-\frac{ -2\sqrt{23}+\sqrt{2}}{ 5\sqrt{5} }\]
jim_thompson5910
  • jim_thompson5910
\[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{-2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{-2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}-2\sqrt{115}}{25}\]
anonymous
  • anonymous
ok I see what I did wrong, I messed up on the second step
jim_thompson5910
  • jim_thompson5910
oops I should have a positive sign
anonymous
  • anonymous
where in front of the 2
jim_thompson5910
  • jim_thompson5910
\[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}+2\sqrt{115}}{25}\]
jim_thompson5910
  • jim_thompson5910
yeah
anonymous
  • anonymous
ok
anonymous
  • anonymous
@jim_thompson5910 i have to go to the bathroom I will be back in about 6 minutes
jim_thompson5910
  • jim_thompson5910
ok
anonymous
  • anonymous
In the meanwhile could you input the information for #5 so I can see how you did it
jim_thompson5910
  • jim_thompson5910
you mean #4 ?
anonymous
  • anonymous
@jim_thompson5910 sorry that took longer than expected
anonymous
  • anonymous
ok so I tried finding the equation for question #4 and #5 in the attachment you sent, but I could not find them
jim_thompson5910
  • jim_thompson5910
alpha is in Q4 270 < alpha < 360 270/2 < alpha/2 < 360/2 135 < alpha/2 < 180 so alpha/2 is in Q2 which means sin(alpha/2) will be positive ------------------------------------------------------- Look on that identity sheet on page 2. Look under the "Half Angle Formulas" section You will use the identity \[\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos(\alpha)}{2}}\] the right side is positive because sin(alpha/2) is positive
anonymous
  • anonymous
ok i see it now
anonymous
  • anonymous
i just plug in and solve, right?
jim_thompson5910
  • jim_thompson5910
yes, \[\Large \cos(\alpha) = \frac{\sqrt{2}}{5}\]
anonymous
  • anonymous
ok so my answer is
jim_thompson5910
  • jim_thompson5910
I'm getting \[\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{5-\sqrt{2}}{10}}\] which is the same thing more or less
anonymous
  • anonymous
ok sweet, alright last problem
anonymous
  • anonymous
its the same thing, but instead of subtracting root, we add this time
anonymous
  • anonymous
oh no, we need a different value
jim_thompson5910
  • jim_thompson5910
you'll use a different identity
jim_thompson5910
  • jim_thompson5910
and you're using beta instead of alpha
anonymous
  • anonymous
yeah ok so I found cosine in terms of beta, which is \[-\frac{ 2\sqrt{5} }{ 5 }\]
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
Is this my equation
jim_thompson5910
  • jim_thompson5910
now simplify that
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
I would rationalize the denominator
jim_thompson5910
  • jim_thompson5910
multiply top and bottom of that inner fraction by sqrt(5)
jim_thompson5910
  • jim_thompson5910
doing that gives \[\Large \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{5-2\sqrt{5}}{10}}\]
anonymous
  • anonymous
Is that as simplified as it gets?
jim_thompson5910
  • jim_thompson5910
yeah
anonymous
  • anonymous
wait I think one more problem got cut out while I was uploading my attachment
anonymous
  • anonymous
Here one sec this is simpler, but it got me confused
jim_thompson5910
  • jim_thompson5910
make a new post. This one is getting very laggy
anonymous
  • anonymous
ok

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