Question

csc(cos^-1 square root 3/2)

Answer 1

you know that sin(arcsin(x))=x i suppose?

Answer 2

\(\color{black}{\csc\left\{\arccos (\sqrt{3}/2)\right\}}\)

this?

Answer 3

second on is right

Answer 4

In general, it goes like this,

\(\color{black}{\csc\left\{\arccos x\right\}}\)

\(\color{black}{\dfrac{1}{\sin\left\{\arccos x\right\}}}\)

Rule: \(\color{black}{\sin \theta =\sqrt{1-\cos^2\theta}}\)
(can be derived from sin²θ+cos²θ=1)

we will apply the rule, except that θ in our case
is that huge arccosine peace.

\(\color{black}{\dfrac{1}{\sqrt{1-\cos^2\left\{\arccos x\right\}
{\color{white}{\Large S}}}}}\)

Rule: cos(arccos θ)=θ
Thus, cos²(arccos θ)=θ²

so lets apply this, we get.

\(\color{black}{\dfrac{1}{\sqrt{1-x^2}}}\)

Answer 5

This is all about knowing the following properties...

Answer 6

(first)

[a] csc(θ) = 1/sin(θ) thus, --> sin(θ)=1/csc(θ)
and sin(θ)•csc(θ)=1

[b] sec(θ) = 1/cos(θ) thus, --> sin(θ)=1/sec(θ)
and cos(θ)•sec(θ)=1

[c] tan(θ) = sin(θ)/cos(θ)
[d] cot(θ) = cos(θ)/sin(θ)
thus, --> tan(θ)=1/cot(θ)
and tan(θ)•cot(θ)=1

☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼
(second, specials)

[e] sin²(θ)+cos²(θ)=1

if you divide everything in [e], by sin²(θ) you get:
1+cot²(θ)=csc²(θ)
thus, we get another property

[f] csc²(θ)-cot²(θ)=1

if you divide everything in [e], by cos²(θ) you get:
tan²(θ)+1=sec²(θ)
thus, we get another property

[g] sec²(θ)-tan²(θ)=1

Answer 7

And at third, the inverse function properties

sin(arcsin (θ))=θ
cos(arccos (θ))=θ
csc(arccsc (θ))=θ
sec(arcsec (θ))=θ
tan(arctan (θ))=θ
cot(arccot (θ))=θ

in other words, for any normal trinogometric function of theta (lets call it G(θ))
G(arcG(θ))=θ

Answer 8

And the playing arounds are also consequences.
Like, I can tell that,
\(\color{black}{\sin \theta =\sqrt{1-\cos^2\theta}}\)

from sin²θ+cos²θ=1

Answer 9

and to make this particular conclusion, I preform these steps:

1. subtract sin²(θ) from both sides
2. take the square root of both sides

Answer 10

lastly, there is a problem with this however, I think, because
really,
\(\color{black}{|\sin \theta| =\sqrt{1-\cos^2\theta}}\)
so what I did is correct only assuming that θ is not negative.

(because √(a²)=|a|, be definition, and so it would be if a is sin(θ)...)