Inverse Trig Functions?

Question

I really don't understand how to do these problems...could someone please help me...

sin (cos^-1 (sqrt 2)/2)

cos (Cos^-1 (3/4))

csc (Tan^-1 (4/5))

Answer 1

you have to be familiar with special angles...

you need to find the angles for the inverse trig function....

1. what is θ such that cos θ = √2 / 2
that is θ = cos^-1 (√2)/2
θ = 45°
sin 45° = √2 / 2

2. the answer is still 3/4 ... property of inverses...

3. let θ = tan^-1 (4/5)
tan θ = 4/5
cot θ = 5/4
cot² θ = 25/16 = csc² θ - 1
csc² θ = 41/16
thus csc θ = √(41)/4

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Answer 2

the answer to the first one is

(sqrt of 2) / 2 or -(sqrt of 2) / 2...

This is because
(cos^-1 (sqrt 2)/2) = 45 deg or pi/4 and 315 deg or 7pi/4, since those are the two with a cos calue of (sqrt of 2)/2. Recall special angles and their trig values.

then, what are the sin values of pi/4 and 7pi/4? again, recall the trig values of special angles.

the second one is

3/4

it's just like asking, what is the COS value of the angle with a COS value of 3/4?

the third one is

(sqrt of 41) / 4

Recall that tan = opposite/adjacent (4/5) in a right triangle. Using the pythagorean theorem, you will determine the hypotenuse to be sqrt of 41. CSC is the reciprocal of sin, therefore get the sin, which is opp/hyp. This will result in 4 / sqrt of 41. Reciprocal is the answer. It has only one value since Tan^-1 have values only in the 1st and 4th quadrant. it has a positive tangent value therefore it is in the 1st quadrant, wherein all trig values are positive...

Answer 3

There are a couple of useful identities you should know.

sin[arccos(u)] = cos[arcsin(u)] = √(1-u²)
sin[arcsin(u)] = cos[arccos(u)] = u
sin[arctan(u)] = cos[arctan(u)] = u/√(1+u²)

With these you get

sin[arccos(√2/2)] = √[1-(√2/2)²] = √(1-½) = √2/2
cos[arccos(3/4)] = 3/4
csc[arctan(4/5)] = 1/sin[arctan(u)] = √[1+(4/5)²]/(4/5) = √41/4

Answer 4

sin (cos^-1 (sqrt 2)/2) ?
Pi/4 =45 degrees
7Pi/4 = 315 degrees
we know that cos( Pi/4) = cos(7Pi/4) = (sqrt 2)/2

cos^-1 (sqrt 2)/2) = Pi/4 or 7Pi/4
sin (cos^-1 (sqrt 2)/2) = sin (Pi/4) or sin (7Pi/4)

answer is +/- (sqrt 2)/2 (depending on which quadrant you are)
Inverse functions are multiple valued functions.
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cos (Cos^-1 (3/4))

cos (Cos^-1 (3/4)) = 3/4

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Answer 5

Question 1
sin 45° = 1 / √2

Question 2
cos 41.4° = 0.75

Question 3
cosec (38.7°) = 1 / sin 38.7° = 1.6