Question

I am struggling with double angle formulas and inverse trig.
How would you rewrite the following as an algebraic expression of x. Cos(2tan^-1(x))

Answer 1

Please present "inverse trig functions" by drawing them in the Draw utility or through Equation Editor. \[\cos (\tan ^{-1}x)\]

Answer 2

This reads: "Find the cosine of the angle whose tangent is x."

I could and will expand that: "Find the cosine of the angle whose tangent is x/1."

Answer 3

Can you draw an angle that has a tangent of x/1? Here the opposite side is represented by x and the adjacent side is represented by 1.

Use the Draw utility, below.

Answer 4

my picture is not wanting to send @mathmale. Sorry if I'm no typing the equations as they're supposed to be I am new to this website.

Answer 5

what's the tangent identity equals to anyway?
I mean hmm \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=?\)

Answer 6

it is equal to OPPOSITE/ADJACENT

Answer 7

ok... so hmmm \(\bf tan^{-1}(x)=\theta \qquad \textit{really means}\qquad tan(\theta)=x\implies tan(\theta)=\cfrac{x}{1}\) maybe?

Answer 8

what... hmm do you think?

Answer 9

Okay what you did right there is you basically interchanged the y and the x to make f'(x) into the original f(x). So does that break the equation down to \[Cos(2(x/1))\]

Answer 10

@jdoe0001

Answer 11

let A be an angle
then you want cos(2A)

A= tan^-1 (x)
so tan A = x/1
draw a triangle, label the opposite and advacent sides (x and 1)
find the hypotenuse
then you can find cos A and sin A
and use those to find cos(2A)

Answer 12

well... x/1 is the tangent value, not the angle itself, the inverse tangent function, is spitting out the angle itself

so..... now that you know who the opposite and trhe adjacent sides are, graph them as mathmale suggested ::)

Answer 13

notice the angle, say

\(\bf tan^{-1}(x)={\color{brown}{ \theta }}\qquad \textit{really means}\qquad tan(\theta)=x\implies tan(\theta)=\cfrac{x}{1}
\\ \quad \\
thus
\\ \quad \\
cos(2tan^{-1}(x))\implies cos(2{\color{brown}{ \theta}})\implies 2cos^2({\color{brown}{ \theta}})-1\implies ?\)

so, if you just graph it, and from there, you'll just need the cosine of it

Answer 14

after finding the hypo. I come up with \[Cos(\cos ^{2}(1\div \sqrt{x ^{2}+1}) - \sin ^{2}(x \div \sqrt{x ^{2}+1}))\] is that correct ? @jdoe0001

Answer 15

you might want to think like this
cos(2A) = cos^2 A - sin^2 A
you know cos A and sin A so you can find cos(2A)

Answer 16

It can get confusing, so I try to keep things simple
I know tan^-1 (x) is some angle, so call it A
from which we get tan(A)= x/1
from that we get sin(A) x/sqr(x^2+1) and cos A 1/sqr(x^2+1)

we want cos(2A) (replace tan^-1 x with the name of the angle A)
cos(2A) by the double angle formula is cos^2 A - sin^2 A
plug in cosA and sinA and we get our answer

Answer 17

not clear ?

Answer 18

when putting into the double angle formula A is x/1 ?

Answer 19

no. The logic is not quite "plugging in" things. We are being tricky, and have to think a bit differently.

Start with this:
if you want
cos(2A) and I tell you cosA and sin A , can you find cos2A ?

Answer 20

ok I see. when you put it into the double angle you use the value from the triangle which we found using \[\tan ^{-1}\] correct ?

Answer 21

yes. You found those correctly.

Answer 22

Ok now that I have those values. how could I turn all my equation into an algebraically expression?

Answer 23

\[ \cos 2( \tan^{-1} x) = \cos(2A) = \left(\cos A\right)^2 - \left(\sin A\right)^2 \]

Answer 24

*\[ \cos(2 \tan^{-1}x) \]

Answer 25

we know \[ \cos A= \frac{1}{\sqrt{x^2+1}}\]
so plot that into the formula
ditto for sin A

Answer 26

you get
\[ \frac{1}{x^2+1}- \frac{x^2}{x^2+1}\]

Answer 27

so technically you just drop the cos and sin and keep the values we found using our triangle ?

Answer 28

I'm not sure what you mean by drop the cosine and sine

we are using the info from the tan(A)= x/1 to find the cosA and the sinA
for this problem, and using those "values" in the double angle formula

Answer 29

we are doing the same thing as, for example,
cos(60)= cos(2*30) = cos^2(30) - sin^2(30)
and we use a calculator (or memorize) the numerical values for cos(30) and sin(30)

In this problem, we don't have numbers, we have "expressions" but it's the same idea.

Answer 30

Ahhh okay its making sense. However when the question asks for an algebraic expression it is telling us to give them an equation to find x without cosine and sine ? so how do we get rid of cosine and sine. to make it an algebraic expression. Maybe I'm over thinking it or approaching it the wrong way.

Answer 31

As you know we can find cos A and sin A as fractions (see above)
we use those values in the double angle formula, and we will get
\[ \cos\left(2 \tan^{-1} x\right) = \frac{1-x^2}{1+x^2} \]

Answer 32

we are not "finding x", we are finding the cos of the mess "in terms of x"

Answer 33

we can test that formula. For example, when x=1
we expect to get 0 (the top 1-x^2 will be 0)

now let's do it step by step:
tan^-1 (1) = 45º
2*45 = 90
cos(90º) = 0
it works!

Answer 34

Okay im still a little confused after we found our CosA and SinA how did we get rid of the sqrt's that came from the hypotenuse ?

Answer 35

do you know the formula for cos(2A) ?

Answer 36

yes its \[\cos ^{2}(A)-\sin ^{2}(A)\]

Answer 37

and the first term is cosine squared
so square the value (or expression if you like) you found for cos A
what do you get ?

Answer 38

or multiply \[ \frac{1}{\sqrt{1+x^2} }\cdot \frac{1}{\sqrt{1+x^2} }\]
top times top and bottom times bottom

Answer 39

Oh okay thats when the square roots get cancelled because squaring a sqrt cancel each other out !

Answer 40

exactly

Answer 41

Okay I see now ! Thank you so much !!!

Answer 42

It is tricky, so you might have to go over the logic and try it on other problems.

Answer 43

Yes it is very Tricky. Would it be the same process when we dont have a double angle inside ?

Answer 44

if it were just cos( tan^-1 x) i.e. cos(A) it's the same idea but easier
if it's cos(A/2) you can use the same idea: find cosA and sinA and use them in the half angle formula

Answer 45

and I suppose you could do cos(3A) as long as you can rewrite that in terms of cosA and sin A

and in all these cases I am assuming A = tan^-1 or sin^-1 or cos^-1
in each case, you can label a triangle with 2 sides and find the 3rd side, and then find the other trig values i.e. sin A cos A or tan A if you needed it.

Answer 46

Okay. but in the situation where we would have for example Sin(Sin^1(x)) The answer would be x correct because it is the composite ?

Answer 47

i meant \[Sin(Sin ^{-1}(x))\]

Answer 48

yes, the answer is immediately x
but we could do it by saying
A= sin^-1 x
sinA = x/1

and we want sin (sin^-1 x) which (replacing sin^-1 x with A ) is
sin A
and of course sin A = x/1 = x

Answer 49

but if we wanted
cos (sin^-1 x)
we would have to do it as above:
let A = sin^-1 x
and so sin A = x/1
and find the other side : sqr( 1-x^2)
and then we can find
cos A

Answer 50

I'm sorry for all the questions I'm just trying to fully understand since my exam is coming up. But in the example ummm... Lets say \[Sin(\cos ^{-1}(x))\]
We would first find the values to our triangle correct ?

Answer 51

which makes \[\cos ^{-1}(x)\] = \[\cos \theta =x \div1 \]

Answer 52

I would write
A= cos^-1 x
then apply cos to both sides to get
cos A = x
write x as a fraction, so x/1

cos A= x/1
now label the sides of the triangle: adj is x , hyp 1 and we find the 3rd side

Answer 53

if our hypotenuse is 1 and we have x as our adj. then our opp would as well be x ?

Answer 54

no, you use pythagoras

Answer 55

a^2 + b^2 = c^2
let b be x and c be 1 and solve for a

Answer 56

which is \[a^2 + b^2 = c^2\]
and since we have hypo. and adj it would be
\[a^2 + x^2 = 1\]

Answer 57

yes

Answer 58

so after making our triangle we come down to our equation sin(x/1) correct ? since we replaced cos^-1

Answer 59

no, you have a "bug" in your thinking
to keep things clear, remember cos^-1 returns an angle. I always rename , for example,
A= cos^-1 x
and change sin(cos^-1 x) to sin(A)
so I know I want the sin of A

Answer 60

okay but at the end how could i rewrite it into an algebraic equation if i simply have Sin(A)

Answer 61

meanwhile, I use
A = cos^-1 x
to write
cosA = x/1
find the 3rd side of the triangle
find sin A = sqr(1-x^2)

and we are done.

Answer 62

ohhhhh So you are just naming the side that we find with our triangle. since its not a double angle

Answer 63

yes.
if the problem was
sin (2 cos^-1 x)
i.e. sin(2A)
we would have to find both sin A and cos A
and use them in the double angle formula to find sin2A

Answer 64

Okay. I see what i was doing ! With double angles we use the formulas and when its a simple angle we just find the measure of the triangle and find the corresponding

Answer 65

yes.

Answer 66

perfect thank you.

Answer 67

yw