SOLUTION: (Please Help...midnight deadline!)I really need to understand this type of problem (I'll give 2 examples):
Simplify the expressions below.
(a) For -00 < t < 00,
cos( tan^-1
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Question 155308: (Please Help...midnight deadline!)I really need to understand this type of problem (I'll give 2 examples):
Simplify the expressions below.
(a) For -00 < t < 00,
cos( tan^-1( t ))=____________
(b) For -1 < w < 1,
cot( cos^-1( w ))+____________
I understand that cos=adj/hyp and tan=opp/adj, but what is the relationship of cos to tan^-1 in this problem? I also understant limiting the range (ie. b), just not the relationship between cot and cos.
Appreciate any help I can get ...thanks.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Simplify the expressions below.
(a) For -00 < t < 00,
cos( tan^-1( t ))=____?________
----------
Consider tan^-1(t)
This is "the angle whose tangent is "t".
Let the angle be theta.
--------------------------
Draw a right triangle.
Let the base be 1 and the height be "t"
Then the angle opposite "t" is theta.
Verify for yourself that tan(theta) = t/1 = t
----------
Using Pythagoras, the hypotenuse = sqrt(1^2 + t^2) = sqrt(1+t^2)
----------
Ans: Then cos(theta) + cos(tan^-1(t)) = t/sqrt(1+t^2)
=============================================================
------------------------------------
(b) For -1 < w < 1,
cot( cos^-1( w ))+_____?_______
--------------------------------------
Let cos^-1(w) = theta
Draw a right triangle.
Let the base be "w".
Then the hypotenuse = 1
Theta is the angle opposite the remaining side.
-------------
Verify for yourself that cos(theta) = w/1 = w
-----------------
Use Pythagoras to find the side opposite theta:
Let that unknown side be "y"
1 = w^2 + y^2
y = sqrt(1-w^2)
Ans: Then cot(theta) = w/[sqrt(1-w^2)]
============================================
Cheers,
Stan H.
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