• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Induction (1 Viewer)

GUSSSSSSSSSSSSS

Active Member
Joined
Aug 20, 2008
Messages
1,102
Location
Turra
Gender
Male
HSC
2009
Prove tan((pi/4)(2n+1))=(-1)^n

gah i keep getting tan(pi/2)'s in my answer
so it fuks it up lol

if anyone can help i thank you in advance lol XDDD
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Have you tried converting the tan(pi/2) into 1's ?
 

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,449
Location
Sydney
Gender
Male
HSC
2009
Prove tan((pi/4)(2n+1))=(-1)^n

n=1
LHS=tan(3pi/4)=-1=RHS
therefore n=1 is true

assume true for some particular n=k, i.e.
tan((pi/4)(2k+1))=(-1)^k

Prove for n=k+1
i.e. tan((pi/4)(2k+3))=(-1)^(k+1)

RHS=-1^k x-1
LHS=-tan (pi-pi(2k+3)/4)=-tan(pi(1-(2k+3)/4)=-tan(pi/4(2k+1))

sub in n=k
LHS=(-1)^kx-1 =RHS
 

GUSSSSSSSSSSSSS

Active Member
Joined
Aug 20, 2008
Messages
1,102
Location
Turra
Gender
Male
HSC
2009
Prove tan((pi/4)(2n+1))=(-1)^n

n=1
LHS=tan(3pi/4)=-1=RHS
therefore n=1 is true

assume true for some particular n=k, i.e.
tan((pi/4)(2k+1))=(-1)^k

Prove for n=k+1
i.e. tan((pi/4)(2k+3))=(-1)^(k+1)

RHS=-1^k x-1
LHS=-tan (pi-pi(2k+3)/4)=-tan(pi(1-(2k+3)/4)=-tan(pi/4(2k+1))

sub in n=k
LHS=(-1)^kx-1 =RHS
sorry i cant quite read your notation lol sorry

but im finding it hard to follow as well
you seem to pull pi's outta nowhere sorry lol
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top