1
The sum of the three smallest positive values of theta such that 4(cos*theta)(sin*theta) = 1 is k*pi. Find k?
↳Redirected from
"How to prove this ?
#2|k^2# then #2|k# for some #k\inZZ#"
#4costhetasintheta = 1#
#=>2costhetasintheta = 1/2#
#=>sin2theta = sin(pi/6)#
#=>2theta=npi+(-1)^n*(pi/6)#
#=>theta=1/2(npi+(-1)^n*(pi/6)) "where "ninZZ#
When #n=0totheta=pi/12#
When #n=1totheta=(5pi)/12#
When #n=2totheta=(13pi)/12#
By the condition of the problem
#pi/12+(5pi)/12+(13pi)/12=kpi#
So #k=19/12#
We can rewrite as
#2costhetasintheta = 1/2#
#sin(2theta) = 1/2#
Now I'm assuming that we're only consider values of #theta# greater than #0#.
#2theta = pi/6, (5pi)/6, (13pi)/6#
#theta = pi/12, (5pi)/12, (13pi)/12#
Now you just have to add the values up
#theta_"sum" = (19pi)/12#
Therefore #k = 19/12#
Hopefully this helps!
Here,
#4costhetasintheta=1#
#=>2sinthetacostheta=1/2#
#=>sin2theta==1/2 > 0=>I^(st) Quadrant orII^(nd)Quadrant#
#=>2theta=pi/6,(pi-pi/6),(2pi+pi/6),(3pi-pi/6),(4pi+pi/6),...#
#=>2theta=pi/6,(5pi)/6,(13pi)/6,(17pi)/6,(25pi)/6,(29pi)/6,...#
#=>theta=pi/12,(5pi)/12,(13pi)/12,(17pi)/12,...#
The sum of the three smallest positive values of theta #=kpi#
So,
#pi/12+(5pi)/12+(13pi)/12=kpi#
#=>(pi+5pi+13pi)/12=kpi#
#=>(19pi)/12=kpi#
#=>19/12=k#
#i.e. k=19/12~~1.5833#
