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anonymous
 one year ago
Need help with some trig problems
anonymous
 one year ago
Need help with some trig problems

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Use the identity cos(A  B) = cos(A)*cos(B) + sin(A)*sin(B)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're given tan(B). Use that to find the value of sin(B)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait we are solving question number 1 right

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2draw a right triangle with the angle beta dw:1438649540480:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2if tan(B) = 1/2, then what does that mean for the right triangle above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin is \[\sqrt{5}/5\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2correct dw:1438649735884:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2what's cos(B) equal to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2\sqrt{5} }{ 5 }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2actually, B is in Q2, so cosine is negative

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2now you do the same for the angle alpha dw:1438649901995:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438649921905:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2sin(alpha) will be negative since alpha is in Q4

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2that will be negative, but otherwise, good

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \cos(\alpha  \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha  \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{2\sqrt{5}}{5}\right) + \left(\frac{\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha  \beta) =??\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[\frac{ \sqrt{2} }{ 5 }\times \frac{ 2\sqrt{5} }{ 5 }+\frac{ \sqrt{23} }{ 5 }\times \frac{\sqrt{5} }{ 5 }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2sin(B) is positive because B is in Q2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I cant get an exact answer on my calculator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \cos(\alpha  \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha  \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{2\sqrt{5}}{5}\right) + \left(\frac{\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha  \beta) = \frac{2\sqrt{2*5}}{5*5} + \frac{\sqrt{23*5}}{5*5}\] \[\Large \cos(\alpha  \beta) = \frac{2\sqrt{10}}{25}  \frac{\sqrt{115}}{25}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Optionally, 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2\sqrt{10}\sqrt{115} }{ 25 }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah that's another way to write the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok how do we do the second one

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2review this list of identities http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look for the "tan(2theta)" identity on the second page

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so you need to find tan(alpha) and plug it into that identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tan is \[\frac{  \sqrt{46}}{ 2}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2alpha is in Q4 where tan is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so can you show me how you plug it in

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \tan(2\alpha) = \frac{2\tan(\alpha)}{1\tan^2(\alpha)}\] \[\Large \tan(2\alpha) = \frac{2\left(\frac{\sqrt{46}}{2}\right)}{1\left(\frac{\sqrt{46}}{2}\right)^2}\] \[\Large \tan(2\alpha) = \frac{\sqrt{46}}{1\frac{46}{4}}\] I'll let you simplify

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i got\[\frac{ 2\sqrt{46} }{ 21 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0great does the attachement you sent me have the equation for the third one as well

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yes, look under "Sum and Difference Formulas" on page 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sin(a)cos(b)+cos(a)sin(b)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0I am having a hard time typing it in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2\sqrt{23}+\sqrt{2}}{ 5\sqrt{5} }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{2\sqrt{5}}{5}\right)\left(\frac{\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}2\sqrt{115}}{25}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok I see what I did wrong, I messed up on the second step

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2oops I should have a positive sign

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where in front of the 2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{2\sqrt{5}}{5}\right)\left(\frac{\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}+2\sqrt{115}}{25}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 i have to go to the bathroom I will be back in about 6 minutes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In the meanwhile could you input the information for #5 so I can see how you did it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you mean #4 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 sorry that took longer than expected

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so I tried finding the equation for question #4 and #5 in the attachment you sent, but I could not find them

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2alpha 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
 one year ago
Best ResponseYou've already chosen the best response.0i just plug in and solve, right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yes, \[\Large \cos(\alpha) = \frac{\sqrt{2}}{5}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I'm getting \[\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{5\sqrt{2}}{10}}\] which is the same thing more or less

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok sweet, alright last problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its the same thing, but instead of subtracting root, we add this time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no, we need a different value

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you'll use a different identity

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2and you're using beta instead of alpha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah ok so I found cosine in terms of beta, which is \[\frac{ 2\sqrt{5} }{ 5 }\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2now simplify that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I would rationalize the denominator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2multiply top and bottom of that inner fraction by sqrt(5)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2doing that gives \[\Large \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{52\sqrt{5}}{10}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that as simplified as it gets?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait I think one more problem got cut out while I was uploading my attachment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here one sec this is simpler, but it got me confused

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2make a new post. This one is getting very laggy
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