1: tan(arcsin(12/13)) and 2: sin(arccos(5/13))?

Question

sorry about all the stupid quetions, but this is a good way for me to study for freshman finals if you guys dont mind lol

Answer 1

The trick to these is to keep in mind SOHCAHTOA, since one way to solve these is through triangles.

Let theta = arcsin(12/13). Take the sin of both sides. Since arcsin and sin are inverses of each other, arcsin should go away, leaving you with

sin(theta) = 12/13

At this point, remember that according to the SOHCAHTOA rules, sin=opp/hyp.

Make a right angle triangle, with the angle as "theta". The "height" of the triangle should be represented as the "opposite", and the bottom of the triangle should be the "adjacent", while the hypotenuse will be the obvious one.

So sin(theta) = 12/13 means you make a right angle triangle with 12 as the opposite and 13 as the hypotenuse. That would make the adjacent equal to sqrt ((13)^2 - 12^2) = sqrt(25) = 5.

Now, here's the trick: Remember what we let theta equal.

cos(theta) = cos (arcsin(12/13)) [which is our original equation and what we want to solve for!]

Using the triangle you made, calculate cos(theta). cos is equal to adjacent over hypotenuse. adjacent, we calculated to be 5, and hypotenuse is 13. Therefore,

cos(theta) = 5/13

and that's our answer.

Try #2 on your own.

Answer 2

well this is the triangle youre dealing with: 5-12-13
it would help to draw a picture
arcsin will give you the angle between the 13 and 5
the tan of that wil be 12/5, because tangent is far side over near side

likewise the second one
the arccos of 5/13 is actually the same angle
the sin of it is 12/13