Need help with some trig problems

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Need help with some trig problems

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Use the identity cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)

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

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