Question

I need help with Pythagorean identity.

Answer 1

if sin(x) = 1/2, what is cos(x) and tan(x)?

Answer 2

\[\sin(x) = \frac{ 1 }{ 2 }\]

on a calculator you can enter in inverse sine

\[sine^{-}(\frac{ 1 }{ 2 }) = x | ArcSin(\frac{ 1 }{ 2 }) = x \]

Answer 3

this will give you the value of the angle theta that when you take sine of that same angle you get 1/2

Answer 4

okay...you lost me at "on a calculator"

Answer 5

i need to know how to do the steps one by one

Answer 6

does cos = √ 3/4

Answer 7

another way would be this

Answer 8

and then tan = sin/cos

Answer 9

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

Answer 10

okay so i plugged in 1/2 but then i don't know my next step

Answer 11

So take a look at the steps below

We use this identity

\[\sin^{2} + \cos^{2} = 1\]

\[(1/2)^{2}+\cos^{2}(x) = 1\]

\[Cos^{2}x = 1-(\frac{ 1 }{ 4 })\]

\[\cos^{}(x) = \sqrt{\frac{ 3 }{ 4 }}\]

Answer 12

so tangent x would be

\[\tan(x) = \frac{ \sin(x) }{ \cos(x) }\]

Answer 13

okay so 1/2 over √ 3/4?

Answer 14

yep let's simplify that

Answer 15

how?

Answer 16

\[Tangent(x) = \frac{ 1 }{ 2*\sqrt{\frac{ 3 }{ 4 } } } = \frac{ 1 }{ 2*\frac{ \sqrt{3} }{ \sqrt{4} } } = \frac{ 1 }{ \cancel\2*\frac{ \sqrt{3} }{ \cancel\2 } } = \frac{ 1 }{ \sqrt{3} }\]

Answer 17

where is the 2 from on the denominator? is it the two from 1/2?

Answer 18

it's just re-written

Answer 19

okay, thank you!

Answer 20

we had

\[\frac{ \frac{ 1 }{ 2 } }{ \sqrt{\frac{ 3 }{ 4 }} } = \frac{ 1 }{ 2*\sqrt{\frac{3 }{ 4 } }}\]

Answer 21

ohhh, okay can you help me another one?

Answer 22

@Photon336