Question

Using a reference triangle, find the exact value of the expression: cos(tan^-1 x^2)
A.1/sqrt x^4+1
B.1/x^4-1
C.x^4/sqrt x^4+1
D.x^4/x^4+1
I answered A

Answer 1

ok noob lemme see your problem

Answer 2

thats my problem above i answered A

Answer 3

I think you mean,
\[\cos(\arctan(x^2))\] right?

Answer 4

yes

Answer 5

Well, there is an intesting formula that relates the cos and the arctan in this case. The question
\[\cos(\arctan(x^2))\]
means: "calculate the cos of the angle whose tangent is equals to x^2"

Answer 6

That means,
\[\tan\alpha=x^2\]And you know
\[1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\]So,
\[\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}\]That means,
\[\cos\alpha=\frac{1}{\sqrt{1+x^2}}\]

Answer 7

yes

Answer 8

Sorry, the last formula should be,
\[\cos\alpha=\frac{1}{\sqrt{1+x^4}}\]And this is the answer.

Answer 9

Well, I tried to do the process, but it seems I also give the answer.

Answer 10

ok thanks

Answer 11

You're welcome.

Answer 12

i have a couple more problems i need help with

Answer 13

Tell me.

Answer 14

here or on the forum

Answer 15

Wherever you want.

Answer 16

ok ill tell here

Answer 17

Identify the middle line for the function: f(t)=1/2sec(t+2)-pi
A.y=-pi
B.y=pi
C.y=pi/2
D.y=-pi/2
I answered C

Answer 18

Let me see.

Answer 19

You mean the function,
\[f(t)=\frac{1}{2\sec(t+2)}-\pi\]?

Answer 20

no

Answer 21

Ok, I see now.

Answer 22

Well, the midline of a trigonometric function I think is another point.

Answer 23

yes

Answer 24

Do you remember the graph of the sine? The midline I think is the line where you measure the amplitude. Or better is the line where the sine is zero.

Answer 25

In this case, is the line where the function with t is zero

Answer 26

So the midline is in -pi

Answer 27

ok

Answer 28

Just the first answer.

Answer 29

so the answer is A?

Answer 30

Yes, the line where the function with t is zero.

Answer 31

ok i have i think 1 more problem

Answer 32

i answered B

Answer 33

Perfect!

Answer 34

ok good thanks

Answer 35

You're welcome.

Answer 36

ok 1 more

Answer 37

i answered A is that also correct or i did it wrong

Answer 38

john?

Answer 39

Ok, you're answer is perfect!

Answer 40

wow really?

Answer 41

ok last problem i promise

Answer 42

Yes, the number that multiplies the t gives you the angular frequency,
\[2\pi f=1\Rightarrow f=1/(2\pi)\] The period is the inverse of the frequency, so,
\[T=1/f=2\pi\]

Answer 43

ok what about this last problem i just sent

Answer 44

Well, what is the value of the y if the function with t goes to zero?

Answer 45

zero?

Answer 46

Exactly.

Answer 47

so would it be y=0?

Answer 48

Yes :)

Answer 49

ok thanks so much!

Answer 50

You're welcome.