Famous Inverse Trig Questions (1 Viewer)

[Slidey]

It's good to know how to do these, since they're within the scope of the course, but typically they would lean towards component B. Id est: Harder questions.

 

[FinalFantasy]

It's good to know how to do these, since they're within the scope of the course, but typically they would lean towards component B. Id est: Harder questions.

deir probably a bit harder but aren't dey in text books as well?

 

[velox]

FF they're almost identical to ones in cambridge/fitz.

How would i do this one?
Show:
arccos (3/11) - arcsin(3/4) = arcsin (19/44)

 

[FinalFantasy]

FF they're almost identical to ones in cambridge/fitz.

How would i do this one?
Show:
arccos (3/11) - arcsin(3/4) = arcsin (19/44)

all of them pretty much the same
just chuck in variables den play around with it
e.g dis one

let A=cos^-1 (3\11) and B=sin^-1 (3\4)
cosA=3\11 and sinB=3\4
sin(A-B)=sinAcosB-cosAsinB=(sqrt(112)\11)(sqrt(7)\4)-(3\11)(3\4)
=28\44-9\44=19\44=sin ( sin^-1 (19\44))
.: A-B=sin^-1 (19\44)
hence "arccos (3/11) - arcsin(3/4) = arcsin (19/44)"

 

[velox]

what about ones like these:

Show:
arctan (3/4) = 2arctan(1/3)

 

[FinalFantasy]

what about ones like these:

Show:
arctan (3/4) = 2arctan(1/3)

show that 3\4=tan (2A) where A=tan^-1 (1\3)

 

[velox]

thanks........

 

[velox]

What about this one? I tried it but ended up getting a completely wrong answer

2arctan2 = pi - arccos (3/5)
Using tan(pi-x) = -tanx

 

[FinalFantasy]

What about this one? I tried it but ended up getting a completely wrong answer

2arctan2 = pi - arccos (3/5)
Using tan(pi-x) = -tanx

u see the RHS has the form pi-x already
so let x=cos^-1 (3\5)
cosx=3\5
let B=tan^-1 (2)
tanB=2

tan (2B)=2tanB\(1-tan²B)=4\(1-4)=-4\3
tan (pi-x)=-tanx=-4\3
.: 2B=pi-x
hence 2tan^-1 (2)=pi-cos^-1 (3\5)

 

[FinalFantasy]

maybe in 3unit u don't have to, but i think u should make arguments that da angles are in first quadrant, otherwise u can't just say their equal

 

[velox]

one thing i dont get, is how does double angle come into this?

 

[Slidey]

Well you agree that pi/2 is an angle? (90 degrees)

Well, arcsin(1) is also an angle! It is pi/2.

All arcfunctions are angles. So when you have two of them - two angles - added together, you can take the sin, cos, tan, whatever of the both angles and apply double angle formulae. And if you want to get it back into a single angle, you can then take the arcsin, etcetera of the your answer so that you have an angle again.

 

[velox]

How about specifically in the below case?

u see the RHS has the form pi-x already
so let x=cos^-1 (3\5)
cosx=3\5
let B=tan^-1 (2)
tanB=2

tan (2B)=2tanB\(1-tan²B)=4\(1-4)=-4\3
tan (pi-x)=-tanx=-4\3
.: 2B=pi-x
hence 2tan^-1 (2)=pi-cos^-1 (3\5)