Question

Please... tan(sin^-1(-4/5))

Answer 1

Let
\[\theta = \sin^{-1} (-\frac{4}{5})\]
Then
\[\ sin \theta = -\frac{4}{5}\]

\[\tan \theta =\pm \frac{4}{x}\]
where
\[x^2 = 5^2 - 4^2 = 3^2\]

Answer 2

what does the negative sign do? in -4/5

Answer 3

also, isn't that the unit circle? why is the hypothesis/radius 5?
I thought it had to be 1

Answer 4

arcsine can only give you a value between -pi/2 and pi/2. So the angle is in the 4th quadrant

Answer 5

when you solve for x using Pythagiran Theorem, you should get x to be positive. I.e, x = +3

so tan(arcsin(-4/5)) = -4/3

Answer 6

-4/3. Please help me on mine.

Answer 7

Hey

Answer 8

@ganeshie8 hey! omg I forgot to ping you

Answer 9

Let's try and do this in another way

Answer 10

Recall that
tanx = sinx/cosx

Answer 11

What do you mean, do you still want my user?

Answer 12

Do you know how to express cosx in terms of sinx ?

Answer 13

how do you do that?

Answer 14

What trig identity do you remember the most ?

Answer 15

"Trig identity" comes up in the next unit, for me.

Answer 16

You must be knowing some

Answer 17

even if you're not supposed to know

Answer 18

1Introduction to Trigonometry
2Trigonometric Functions
3 Working with Trigonometric Functions
4 Trig Identities

I'm on unit 3 right now.

Answer 19

Well. yeah. Are you talking about simple functions like sin/cos/tan/csc etc

Answer 20

I don't know how to 'express cosx in terms of sinx'

Answer 21

sin^2x + cos^2x = 1


have seen this identity before ?

Answer 22

Pythagorean!

Answer 23

Yes, if you're happy with that identity, you may use it to express cosx in terms of sinx

Answer 24

\[\cos(x) = \sqrt{1-\sin^2(x)}\]

Answer 25

Yep,
So what's tanx in terms of sinx only ?

Answer 26

\[tanx =\frac{ \sin (x) }{ \sqrt{1-\sin^2(x)} }\]

Answer 27

No?

Answer 28

Yep

Answer 29

use that formula to simplify the given expression

Answer 30

Replace x by sin^-1(-4/5)

Answer 31

Are we allowed to do that? It's just "sin(x)", not "sin^-1(x)"...

Answer 32

You're allowed do anything you want as long as you do it to both sides and follow rules. So go-ahead

Answer 33

\[\tan(x) = \frac{ \sin(\sin^-1(-4/5)) }{ \sqrt{1-\sin^2(\sin^-1(-4/5))} }\]

Answer 34

? Seems more complicated..

Answer 35

You forgot to replace x in the lefylt hand side

Answer 36

\[\tan(\sin^-1(-4/5))) = \frac{ \sin(\sin^-1(-4/5)) }{ \sqrt{1-\sin^2(\sin^-1(-4/5))} }\]

Answer 37

It only looms complicated, but its really easy. sin^-1 eats sin

Answer 38

"eats"?

Answer 39

Yes, An inverse function undoes its function

Answer 40

Let me ask you a question

Answer 41

Suppose you took 10$ from your friend,
To undo your debt, what must you do ?

Answer 42

Give back $10

Answer 43

Yes, so if take(x) is your function,
then its inverse function is give(x)

Answer 44

what's the value of

give(take(10)) ?

Answer 45

@ganeshie8 the value from whose POV? ;P

@dumbcow's method works too, except sin^-1 only returns angles in quadrants 1 and 4, so the angle is only quadrant 4 (not in 3)

Answer 46

Yeah I should be more clear on that question :
Suppose you had $100 in your pocket(x = 100).

Then you took $10 :
take(100) = 110

Then you gave back $10 :
gave(110) = 100

Answer 47

The net effect of taking and then giving is nothing.

It brings you back to your original state :
gave(take(100)) = 100

Answer 48

That works better. The one you originally described give(take(10)) = 0 (for your own POV)

Answer 49

im sorry i had to leave yesterday.. wait.. im reading this now

Answer 50

okay.. so what happens next?

Answer 51

\[\tan(\sin^-1(-4/5))) = \frac{ \sin(\sin^-1(-4/5)) }{ \sqrt{1-\sin^2(\sin^-1(-4/5))} }\]

Answer 52

sin eats sin^-1
simply cancel them

Answer 53

\[\tan(\sin^-1(-4/5))) = \frac{ 1 }{ \sqrt{1-\sin} }\]

????

Answer 54

doesn't look right

Answer 55

The sin cancels out inverse sin. It does NOT cancel out the number inside the sin.

Answer 56

Also keeping mind,

sin^2(x)
is just another way to write
(sin(x))^2

Answer 57

^Yeah don't forget that either

\[\tan(\sin^-1(-4/5))) = \frac{ \sin(\sin^-1(-4/5)) }{ \sqrt{1-\left( \sin(\sin^-1(-4/5)) \right)^2} }\]
I still think @dumbcow's method is the easier one, though

Answer 58

Yes, this is supposed to be an alternative

Answer 59

What's dumbcow's method?

Answer 60

The method we used in a question you did yesterday, @elusive... drawing a triangle.

Answer 61

I think you have almost finished the problem
Cancel the sin, sin^-1 and see if you can simplify

Answer 62

The part I don't understand is.. Why are the triangles in quadrant 3 and 4? Because the sign is negative?

Answer 63

\[\tan(\sin^-1(-4/5))=\frac{ -4/5 }{ 1- -4/5^2}\]

Answer 64

Yes, sin is negative in 3, 4 quadrants.
But this is only half of the answer

Answer 65

The other half had to do with domain of sin^-1

Answer 66

Looks you forgot the radical in the bottom

Answer 67

\[\tan(\sin^-1(-4/5))=\frac{ -4/5 }{ 41/25}\]

how do you simplify further, now?

Answer 68

You forgot to square it... in the denominator, a negative number squared is positive

Answer 69

\[\tan(\sin^-1(-4/5))=\frac{ -4/5 }{ 9/25}\]

Answer 70

\[\dfrac{-4/5}{\sqrt{1-9/25}}\]

Answer 71

I would also like to know the triangle method.. I need to memorize some rules and properties?

Answer 72

@elusive the very first post in this question shows the triangle method. And you and I used it in another question of yours.

Answer 73

Not really, but you need to know some more about inverse functions first.

Answer 74

thank you, but how do you know which quadrant to put the triangles in??

Answer 75

You have to know/use the domains of the inverse functions to decide

Answer 76

Looking at sinx = -4/5, you figure out that x must be somewhere in 3rd or 4th quadrants

Answer 77

when graphing the inverse trig function, you just simply switch the x and y.. right? i think that's easy. I don't know what this thing we're doing right now, though.

Answer 78

That's the first step

Answer 79

The domains and range simply switch, right?

Answer 80

How do you know an inverse exists ?

Answer 81

humans create them

Answer 82

Heard of horizontal line test ?

Answer 83

yup

Answer 84

Yes, we create them and we want them to be well defined