TRIG HELP!!!:'(

[tashadewan]

im having real struggles with this new unit of trig in calc...its only been 3 days and i already have a test tomm....i cant solve this hmwk question

Write cos 3A in terms of cos A....i tried but i always wind up wit some sin in it n i dont know wht to do wit it.

the second one im having a terrible time wit is tan(x+y)tan(x-y)=[sin^2x-sin^2y]/[cos^2x-sin^2y]
i cant prove this...im stuck...theres a difference of squares in it but i cant figure out where to go from tehre

the last question is...how do i find the critical number for s=2sint+sin2t.
I found the derivative to be 2cos t +2cos 2t but i cant solve for 0 coz i dont know how to.....

 

[ChaoticLlama]

Set the derivative equal to 0
2cos t +2cos 2t = 0

2cost = -2cos2t
cost = -cos2t

now think, for what value of t will cost be equal to -cos2t.

t = pi/3 is a critical number

 

[tashadewan]

i dont get where u got the 60 degrees from...trial and error??

 

[askmemath]

cos3A = cos(2A+A)

cos2acosa - sin2asina

(2cos^2a-1)cosa - (2sinacosa)sina

2cos^3a -cos a - 2sin^2acosa

sin^2a= 1-cos^2a

I hope you can take it from here.You have to simply substitute and solve.

 

[askmemath]

tan(x+y)tan(x-y)=[sin^2x-sin^2y]/[cos^2x-sin^2y]

[math] \tan(x+y)=\frac {sin(x+y)} {cos(x+y)} \frac {sinxcosy+cosxsiny} {cosxcosy - sinxsiny} similarly for tan(x-y) YOu should get (a+b)(a-b) which = \{a^2}- \{b^2} [\math] Which is your final answer[/math]