more questions :) (1 Viewer)

[currysauce]

determine

sec^-1 (2)

hehe i dunno...

also show that 2tan^-1(2) = pi-cos^-1(3/5)

 

[Trev]

1.
y=sec^-12
sec(y)=2
Since sec(y)=1/cos(y) then
1/cos(y)=2
cos(y)=1/2
y=cos^-1(1/2) = pi/3 (since y=sec^-12).

 

[Trev]

2.
Draw a triangle to work out that, cos^-1(3/5) is equal to tan^-1(4/3).
So it becomes:
2tan^-1(2)+tan^-1(4/3)=pi
Let tan^-1(2)=A and tan^-1(4/3)=B
2A+B=pi
Take the tan of both sides
tan(2A+B)=0
(tan2A+tanB)/(1-tan2AtanB)=0
tan2A+tanB=0
(2tanA)/(1-tan²A)+tanB=0
Sub in A and B to show that the LHS is equal to zero:
LHS=[2tan(tan^-12)]/[1-tan²(tan^-12)]+tan[tan^-14/3]
[2.2]/[1-2²]+4/3 = -4/3+4/3 = 0 = RHS.
There is probably a shorter way but I can't remember.

 

[currysauce]

thankyou very much

 

[who_loves_maths]

take the tan of both sides of 2arctan(2) = pi - arccos(3/5) then use tan identities to show LHS = RHS is also an expedient method.