anonymous
  • anonymous
csc(cos^-1 square root 3/2)
Trigonometry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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idku
  • idku
you know that sin(arcsin(x))=x i suppose?
idku
  • idku
\(\color{black}{\csc\left\{\arccos (\sqrt{3}/2)\right\}}\) this?
anonymous
  • anonymous
second on is right

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idku
  • idku
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. \(\color{black}{\dfrac{1}{\sqrt{1-x^2}}}\)
idku
  • idku
This is all about knowing the following properties...
idku
  • idku
(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
idku
  • idku
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
idku
  • idku
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
idku
  • idku
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
idku
  • idku
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)...)

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