inverse trig struggle (1 Viewer)

[RideTheLightnin]

hey guys, im just unsure about this question
its from fitzy on page 146 question 4d

it reads: Show that

*=that inverse -1 thing

cos* 3/5 - tan* (-3/4) = pi/2

------------------------
so what i do is i make it

cos* 3/5 + tan* 3/4 = pi/2

then i draw two triangles which both give me the same angle of around 37deg, but i have no idea what im supposed to be doing next...

any help is MUCH appreciated
cheers

 

[insert-username]

I would simply plug the values into the calculator, since that bunch of questions doesn't say "don't use a calculator". That's the easiest way to do it.


I_F

 

[Riviet]

cos^-1(3/5) refers to angle opposite the side length of 4 in the right triangle. tan^-1(3/4) refers to angle opposite the side length of 3 in the triangle. But since angle sum of a triangle is pi and we know the third angle is a right angle [ie is pi/2], then cos^-1(3/5) + tan^-1(3/4) = pi/2

 

[RideTheLightnin]

oh ok thanks mate......it seems so obvious now lol
but i just wanted to make sure....god i hate cramming lol
but anyway cheers

 

[rama_v]

oh ok thanks mate......it seems so obvious now lol
but i just wanted to make sure....god i hate cramming lol
but anyway cheers

If you want to do it more algebraically:
cos^-1 3/5 - tan^-1 (-3/4) = pi/2

Let x = cos^-1 3/5 and y = tan^-1 (-3/4)

cos x = (3/5) and tan y = (-3/4)

sin x = (4/5) , sin y = (-3/5), cos y = (4/5)

x - y = pi/2
so sin(x-y) = sin(pi/2)
LHS = sin(x-y) = sinxcosy - cosxsiny
= (4/5)(4/5) - (3/5)(-3/5)
= 16/25 - (-9/25)
= 25/25
= 1
= sin(pi/2) = RHS

 

[davin]

show that means a caclulator is out...that doesn't actually show anything

i'm gonna use the notation arc for inverses.... (inverse sin is arcsin, inverse cos is arccos, etc)

arccos( 3/5) - arctan(-3/4) = pi/2

for reference, we can see that the cosine for the first angle [arccos(3/5)] is 3/5, and that then the sine is either + or - 4/5 (using those triangles you'd have made)
similarly, arctan(-3/4) has a cosine of (4/5) and a sine of (-3/5) [we know the sine has the negative because cosine has to be positive since our range for arctan is between -pi/2 and pi/2]
we can then take the sine or cosine of both sides using the difference of sines or cosines on the right side, in this case, i'm picking cosine, which gives:
cos{arcos(3/5}cos{arctan(-3/4)} + sin{arccos(3/5)}sin{arctan(-3/4)} = cos(pi/2)
putting the values we defined earlier into this, we'll get
(3/5)(4/5) + (4/5)sin{arctan(-3/4)} = 0
(12/25) + (4/5)sin{arctan(-3/4)} = 0 simplifying, and solving for sin{arctan(-3/4)}
(4/5)sin{arctan(-3/4)} = -12/25
sin{arctan(-3/4)} = (-3/5)
taking the sine inverse of both sides:
arctan(-3/4) = arcsin(-3/5)
which a triangle can show...or maybe someone has a more difinitive proof for that last statement

 

[insert-username]

show that means a caclulator is out...that doesn't actually show anything

Isn't that only the case if you have pronumerals instead of values? In this case, you have values, so you can evaluate the two and show they add to pi/2. That's what I've been taught, anyway.


I_F

 

[rama_v]

Isn't that only the case if you have pronumerals instead of values? In this case, you have values, so you can evaluate the two and show they add to pi/2. That's what I've been taught, anyway.


I_F

I don't think so... For example it would not be reasonable to evaluate limits for an expression as x-> infinity simply by plugging in a large number into the expression. In this case how will you be able to show that the two numbers add to pi/2? pi/2 is an infinitely long irrational number..plugging in numbers is not mathematically valid for this example.