inverse trig question (1 Viewer)

[i-insomnia]

Show that:
tan^-1 (4) - tan^1 (3/5) = pi/4

sin^-1 (3/5) + tan^-1 (7/24) = cos^1 (3/5)

thanks.

 

[Drongoski]

Show that:
tan^-1 (4) - tan^1 (3/5) = pi/4


thanks.

tan(LHS) = [ tan(tan^-1 {4}) - tan(tan^-1 {3/5})]/[1 + tan(tan^-1{4} )tan(tan^-1{3/5})

= [4 - 3/5 ]/[1 + 4 x (3/5) ] = [17/5] / [17/5] = 1

.: LHS = pi/4 [since its tangent = 1]

 

[i-insomnia]

tan(LHS) = [ tan(tan^-1 {4}) - tan(tan^-1 {3/5})]/[1 + tan(tan^-1{4} )tan(tan^-1{3/5})

sorry how did you get this step?

 

[Drongoski]

sorry how did you get this step?

Remember tan^-1 4 and tan^-1 (3/5) are angles (like the A and B above)

 

[Drongoski]

Show that:

sin^-1 (3/5) + tan^-1 (7/24) = cos^-1 (3/5)

thanks.

This question is a little trickier.


Remember each expression is an angle: say they are A, B, C respectively. Construct a right-angled triangle with sides: 3, 4 and 5 [easier to see what's happening this way] The angle opposite the side 3 is A, the one opp 4 is C.





Edit: omitted the last "tan" earlier.

 

[bored of fail 1]

let A=sin^-1(3/5) -> sinA = 3/5 -> cos(A) = 4/5

B = tan^-1 ( 7/24 ) -> tan(B) = 7/24 -> sin(B) = 7/25 -> cos(B) = 24/25


now cos ( A +B) = cos(A)cos(B) - sin(A)sin(B) = (4/5)(24/25) - (3/5)(7/25) = 3/5

therefore take inverse cos both sides and result is shown

dont know what drongskis doing lol

 

[bored of fail 1]

yeh thats a really complicated way of doing it drongoski lol

 

[bored of fail 1]

This question is a little trickier.


Remember each expression is an angle: say they are A, B, C respectively. Construct a right-angled triangle with sides: 3, 4 and 5 [easier to see what's happening this way] The angle opposite the side 3 is A, the one opp 4 is C.

lol i have idea how u got the last line 4/3 = cos^-1 3/5

 

[jyu]

lol i have idea how u got the last line 4/3 = cos^-1 3/5

meant to be tan^-1 4/3 = cos^-1 3/5

 

[Drongoski]

yeh thats a really complicated way of doing it drongoski lol

You are right; your way is better. Good work.

 

[i-insomnia]

thanks.

 

[i-insomnia]

do you guys mind if you help me out a bit more? ..
show that:

sin^-1(x) = cos^ -1 sqrt(1-x^2) , 0 <= x <= 1

thank you so much

 

[bored of fail 1]

its just pythagorus

 

[bored of fail 1]

let A= sin^-1(x)

so sin(A) = x

now we can rerite what we need to prove as sin^-1(x) = cos^-1(sqrt( 1- sin(A)^2 ) )

now the RHS = cos^-1( sqrt (cos(A) )^2 ) = cos^-1(cos(A) = A = sin^-1(x) = LHS

therefore proven

 

[Drongoski]

thanks.

To whom??