trignometry questions, Exams of Mathematics

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2021/2022

Uploaded on 04/23/2023

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Quiz 10 - Math 1022 Name:
Directions: Complete these problems on this paper or separate paper and show all work. Afterwards,
take a picture of your solutions and convert the pictures into a single .pdf file. You can use an app such
as Adobescan, camscanner, or iphone notes to do this easily. Go to Canvas Assignments and find this
assignment and upload your .pdf file there and submit it.
1. Compute arctan(tan(5π/4))
We have 5π
4is outside the range of arctan, which is (π/2, π/2). However, we know
tan 5π
4= 1 = tan π
4
Since π
4is in the range of arctan, we have that
arctan(tan(5π/4)) = π/4.
2. Compute sin(arctan(2/3))
Set θ= arctan(2/3). Then θlies in a right triangle with opposite 2 and adjacent 3, thus hypotenuse
equal to 4 + 9 = 13. So
sin(arctan(2/3)) = sin(θ) = 2
13.
3. Simplify tan(arcsin(x/4)).
Set θ= arcsin(x/4). Then θis the interior angle of a right triangle with opposite x, hypotenuse 4,
and adjacent 16 x2. Thus
tan(arcsin(x/4)) = tan(θ) = x
16 x2.
4. Simplify cos(2 arccos(x))
Set θ= arccos(x). Then θis the interior angle of a right triangle with adjacent x, hypotenuse 1, and
opposite 1x2. We have
cos(2 arccos(x)) = cos(2θ) = cos2(θ)sin2(θ) = x2
11x2
1= 2x21.
1

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Quiz 10 - Math 1022 Name:

Directions: Complete these problems on this paper or separate paper and show all work. Afterwards,

take a picture of your solutions and convert the pictures into a single .pdf file. You can use an app such

as Adobescan, camscanner, or iphone notes to do this easily. Go to Canvas → Assignments and find this

assignment and upload your .pdf file there and submit it.

  1. Compute arctan(tan(5π/4))

We have

5 π

4

is outside the range of arctan, which is (−π/ 2 , π/2). However, we know

tan

5 π

4

= 1 = tan

π

4

Since π 4 is in the range of arctan, we have that

arctan(tan(5π/4)) = π/ 4.

  1. Compute sin(arctan(2/3))

Set θ = arctan(2/3). Then θ lies in a right triangle with opposite 2 and adjacent 3, thus hypotenuse equal to

  1. So

sin(arctan(2/3)) = sin(θ) =

  1. Simplify tan(arcsin(x/4)).

Set θ = arcsin(x/4). Then θ is the interior angle of a right triangle with opposite x, hypotenuse 4, and adjacent

16 − x^2. Thus

tan(arcsin(x/4)) = tan(θ) =

x √ 16 − x^2

  1. Simplify cos(2 arccos(x))

Set θ = arccos(x). Then θ is the interior angle of a right triangle with adjacent x, hypotenuse 1, and opposite

1 − x^2. We have

cos(2 arccos(x)) = cos(2θ) = cos 2 (θ) − sin 2 (θ) =

x^2

1

1 − x^2

1

= 2x 2 − 1.