anonymous
  • anonymous
Trigonometry..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
-dies because my head hurts and I'm only in Geometry-
anonymous
  • anonymous
@abb0t @swami @sophisticated.CHIMP @Psymon @taylor82 @calculusxy

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More answers

anonymous
  • anonymous
k..Ta anyways!!
anonymous
  • anonymous
@Falco276 @Hero @karatechopper @april115 @ashley_f97 @zepdrix @litchlani @Loser66
anonymous
  • anonymous
@Luigi0210
Luigi0210
  • Luigi0210
@Loser66 any ideas?
anonymous
  • anonymous
Whew!! I th8 I was the only one baffled!!
Loser66
  • Loser66
hand off.!! I don't see any outlet.
dan815
  • dan815
|dw:1377906912031:dw|
dan815
  • dan815
|dw:1377906994988:dw|
dan815
  • dan815
or use cos2theta trig identity
dan815
  • dan815
|dw:1377907112177:dw|
dan815
  • dan815
|dw:1377907222029:dw|
anonymous
  • anonymous
yup.....matches with z key !!!
dan815
  • dan815
|dw:1377907258936:dw|
dan815
  • dan815
then u can find what dsin2theta is similary
dan815
  • dan815
b*sin2theta
dan815
  • dan815
do u want me to solve 3 too?
anonymous
  • anonymous
and the third one???
dan815
  • dan815
ok
anonymous
  • anonymous
yes kind of !!
anonymous
  • anonymous
wat is the overall answer for 2nd one???
dan815
  • dan815
youd have to complete that simplification then
dan815
  • dan815
im sure that b stuff with simiplify nicely too
dan815
  • dan815
|dw:1377907455094:dw|
anonymous
  • anonymous
overll answer is a!!
dan815
  • dan815
oh 1 sec i didnt subtract
dan815
  • dan815
|dw:1377907741892:dw|
dan815
  • dan815
|dw:1377907821720:dw|
dan815
  • dan815
|dw:1377907908947:dw|
dan815
  • dan815
see 2 = a
dan815
  • dan815
now moving on to 3
dan815
  • dan815
gonna have to try a couple trig identities to see which one works out for 3
anonymous
  • anonymous
whew!!*cleans sweat on forehead*!!
dan815
  • dan815
|dw:1377908097819:dw|
dan815
  • dan815
hmm
dan815
  • dan815
|dw:1377908308404:dw|
dan815
  • dan815
that trig identity will help
dan815
  • dan815
|dw:1377908341982:dw|
dan815
  • dan815
|dw:1377908524595:dw|
dan815
  • dan815
okay now!
dan815
  • dan815
we are kind of reduced back to the stuff we did in question 2
dan815
  • dan815
ill let you try stuff on your own from here
dan815
  • dan815
i think you can get an expression for sin3x+sin3y if you break that up into trig identities again, where you can use the fact that sin(x-y/2)=1/sqrt(root3+1)^2
dan815
  • dan815
just mess around with it you can do it!!
dan815
  • dan815
use this trig sheet if you get stuck http://www.maths.manchester.ac.uk/~cds/internal/tables/trig.pdf
anonymous
  • anonymous
I get only one equation ..........how can I solve them??
dan815
  • dan815
hooolld on!
dan815
  • dan815
umm sin(-angle) =?
dan815
  • dan815
-sin oh crap
dan815
  • dan815
this is + then
dan815
  • dan815
|dw:1377908762855:dw|
dan815
  • dan815
|dw:1377908773402:dw|
dan815
  • dan815
cause of that
dan815
  • dan815
sorry but i g2g right now!! ill help you solve this if u still arnet done by the time i get back
anonymous
  • anonymous
k.....me too gtg.
anonymous
  • anonymous
@dan815 hey cntnu??
anonymous
  • anonymous
can u @dan815 ??:)
anonymous
  • anonymous
@Jack1
Jack1
  • Jack1
hey, still readin the novel above, standby ;)
anonymous
  • anonymous
3rd question
anonymous
  • anonymous
a'ight
Jack1
  • Jack1
k, i'm just fumbling here, so bare with me, but picking up where dan left off: |dw:1377974752576:dw|
anonymous
  • anonymous
kay
Jack1
  • Jack1
and using the trig ident : sin 3A = 3* sin (A) - 4* sin^2 (3 A) also equals we get : sin x + sin y = sqrt 3 (cos y - cos x) which has LHS transform of: sin x + sin y => sin 3A = 3* sin (x) - 4* sin^3 (x) + 3* sin (y) - 4* sin^3 (y)
anonymous
  • anonymous
k
Jack1
  • Jack1
so if you check out the site dan posted earlier: http://www.maths.manchester.ac.uk/~cds/internal/tables/trig.pdf about halfway down the page things start becoming a bit clearer: cos A - cos B = -2 * sin [(A+B)/2] * sin [(A-B)/2] to be clear; in our case though its more like: cos B - cos A = -2 * sin [(B+A)/2] * sin [(B-A)/2] aka cos y - cos x = -2 * sin [(y+x)/2] * sin [(y-x)/2] so back to our previous point: sin 3A = 3* sin (A) - 4* sin^2 (A) which also equals: sin 3A = 2 * cos [(3 A)/2] * sin [(3 A)/2] so sin x + sin y => sin 3A = 3* sin (x) - 4* sin^3 (x) + 3* sin (y) - 4* sin^3 (y) => 2 * cos [3x/2] * sin [3x/2] + 2 * cos [3y/2] * sin [3y/2]
anonymous
  • anonymous
hmm

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