Question

Calculate sin(arctan2-arcsin(1/sqrt(10))),
(No calculator).

Answer 1

I'm initially thinking of using sin(u-v)=sinucosv-sinvcosu, which leads me to:
sin(arctan2)*cos(arcsin(1/sqrt10))-cos(arctan2)/sqrt10
Quite stuck here, is there any other path that is easier, is there something someone sees that I'm missing? :)

Answer 2

Draw some triangles

Answer 3

I don't follow. :(

Answer 4

let \[\Large \theta = \arcsin \frac{ 1 }{ \sqrt{10} }\] then make the triangle above

Answer 5

I follow for that; I don't see how it applies to calculating our main thing though. :/

Answer 6

\[\sin(\arctan(2)-\arcsin(\frac{1}{\sqrt{10}}))\]

Answer 7

It simplifies the arcsin, you can now find cosine. (use the same triangle)

cos(arcsin(1/sqrt10)) can be simplified.

Answer 8

How?

Answer 9

I think you were on the right track with this:
sin(arctan2)*cos(arcsin(1/sqrt10))-cos(arctan2)/sqrt10

Answer 10

cos(arcsin(1/sqrt10))

remember that...
\[\Large \theta = \arcsin \frac{ 1 }{ \sqrt{10} }\]

so you want cos theta.

Answer 11

\[\Large \cos(\arcsin(1/\sqrt10)) = \cos \theta\] since theta is = arcsin(1/sqrt10)

Answer 12

You can do the same with arctan2. Make a triangle, find the other sides... let arctan2=phi, then you need sin phi, since you have a sin(arctan2)...

sin(arctan2)*cos(arcsin(1/sqrt10))-cos(arctan2)/sqrt10

Answer 13

Oh and you need cos phi, since there's a cos(arctan2)

Answer 14

I see! So you're going back to the triangle after you've switched them in?

Answer 15

I think I understand what you mean :)

Answer 16

Yes, that's really the only way to do these... make triangles to get rid of inverses... like sin(arccosx) can be simplified because arccosx = some angle. So it's sin(some angle).

The triangles just help you find all the other sides you need for tan, sin, cos of that angle. Or secant, cosecant, cotangent too.

Answer 17

We never utilize secants etc in Swedish math I think.
Really like your way of solving this! Man I love these calculations, so simple yet so deep! :D

Answer 18

(In our undergrad courses that is)

Answer 19

haha yes these are fun. But they are pretty confusing at first.. and even sometimes later, i'll think "wait, what the hell was i doing again...?"

Answer 20

Indeed! Haha

Answer 21

But if you simplify them bit by bit they're easier, and make sure to name your angles... eg let phi =arctan2 and so on. That way you aren't remember what's what in your head.

Answer 22

Did you solve for the answer? It's beautiful as well! :D

Answer 23

looks like sin(arctan2-arcsin(1/sqrt(10))) = (sqrt2)/2

which means... arctan2-arcsin(1/sqrt(10)) = 45 degrees.

Answer 24

;)