Answer:
Let
x=sin-¹u
Sinx=u
let y=tan-¹v
tany=v
Substituting
Sin[x + y]
Applying the sine expansion
Sinxcosy + CosxSiny
Recall x =Sin-¹u
y=tan-¹v
Sin(Sin-¹u)Cos(tan-¹v) +Cos(sin-¹u)Sin(tan-¹v)
Now at this point
Here's what you do
For the first expression
Sin(Sin-¹u)
Let's simplify this
Let P = Sin-¹u
Taking sine of both sides
SinP=u
Draw a Right angled angle for this
Since Sine from SOHCAHTOA is OPP/HYP
Where P is the angle and u is the opposite and 1 is the hypotenuse since u is the same as u/1
substituting Sin-¹u = P
You have
Sin(Sin-¹u) = SinP
and from the triangle you drew
SinP = u
Taking the second express
Cos(tan-¹v)
Let Q=Tan-¹v
taking tan of both sides
tanQ=v
Draw a right angled triangle for this too
Since Tan from SOHCAHTOA is OPP/ADJ
Find the Hypotenuse cos you'll need it
Now Let's do the substitution again
We first said tan-¹v = Q
When we substitute it in Cos(tan-¹v)
We have CosQ
Cos Q from the second right angle triangle you drew is 1/√1+v²
Because CAH is adj/Hyp
So
the first part of the original Express
Which is
Sin(Sin-¹u)Cos(tan-¹v) is now simplified to
u(1/√1+v²).
Let's Move to the second part of the Original Expression
Cos(Sin-¹u)Sin(tan-¹v)
From our first solution
We said Sin-¹u= P
So replacing it here
we have Cos(sin-¹u) = CosP
let's leave the second one for now which is sin(tan-1v) We'll deal with this after the first
so Cos(Sin-¹u) = CosP
we can still use our first Right angle triangle for this because the angle was P.
so Cos P from that triangle will be
CosP= √1-u²
Now onto the next
Sin(tan-¹v)
From the Second solution of the first we did
we said let Tan-¹v =Q
Substituting this
we have
Sin(tan-¹v) = SinQ
using the second Right angle triangle because its angle is Q
We have
SinQ= v/√1+v²
Answer for second phase Which is
Cos(sin-¹u)Sin(tan-¹v) = √1-u²(v/√1+v²)
We're done
compiling our answers
The answer to
Sin(Sin-¹u - tan-¹v) = u(1/√1+v²) + [(√1-u²)(v/√1+v²)]
You can still choose to factor out 1/√1+v² since it appears on both sides
