Question

An inverse trig-function, inside a (reg) trig-function. (Examples - I will be adding examples from time to time)

Answer 1

\({\bbox[5pt, #99ffcc ,border:2px solid black ]{ \bf Example~~\text{#}1 }}\)

\(\color{#000000 }{ \displaystyle \sin\left(\arccos \sqrt{1/4}\right) }\)

We know that:
\(\color{#000000 }{ \displaystyle \sin^2\theta+\cos^2\theta=1 }\)
\(\color{#000000 }{ \displaystyle \sin^2\theta=1-\cos^2\theta }\)
\(\color{#000000 }{ \displaystyle \sqrt{\sin^2\theta}=\sqrt{1-\cos^2\theta} }\)
\(\color{#0000ff }{ \displaystyle \sin\theta=\sqrt{1-\cos^2\theta} }\)

In our case, \(\theta\) is \(\color{#000000 }{ \displaystyle\left(\arccos \sqrt{1/4}\right) }\). So will apply this,

\(\color{#000000 }{ \displaystyle \sqrt{1-\cos^2\left(\arccos \sqrt{1/4}\right)} }\)

Knowing that

\(\color{#000000 }{ \displaystyle \cos\left(\arccos x\right)=x }\)
thus, \(\color{#000000 }{ \displaystyle \cos^2\left(\arccos x\right)=x^2 }\)
and so is any, \(\color{#000000 }{ \displaystyle \cos^n\left(\arccos x\right)=x^n }\)

So, we have,

\(\color{#000000 }{ \displaystyle \sqrt{1-\left[ \sqrt{1/4}\right]^2} }\)
\(\color{#000000 }{ \displaystyle \sqrt{1-(1/4)} }\)
\(\color{#000000 }{ \displaystyle \sqrt{3/4} =\sqrt{3}/2}\)

Answer 2

\(\color{#000000 }{ \displaystyle \sec\left(\arccos \frac{1}{2}\right) }\)

since \(\sec\theta = 1/\cos\theta\),

\(\color{#000000 }{ \displaystyle \frac{1}{\cos\left(\arccos \frac{1}{2}\right) } }\)

We apply the same rule, \(\cos(\arccos x)=x\), and we therefore get:

\(\color{#000000 }{ \displaystyle \frac{1}{~\dfrac{1}{2}~} \Rightarrow 1\div \frac{1}{2}\Rightarrow1\times \frac{2}{1}\Rightarrow\color{blue}{2} }\)

(I will start coloring the answer in blue for convinience)

Answer 3

\(\Huge\color{white}{yay}\)

Answer 4

*claps* *sending appalade* xD

Answer 5

\(\color{#000000 }{ \displaystyle \tan(\arcsin 0.5 ) }\)

since \(\tan \theta = \sin \theta / \cos\theta \)

\(\color{#000000 }{ \displaystyle \frac{\sin(\arcsin 0.5 )}{\cos(\arcsin 0.5 )} }\)

the top comes out right away... Another important rule
(of same nature as this rule by the cosine function)
\(\sin(\arcsin x )=x\)

\(\color{#000000 }{ \displaystyle \frac{0.5}{\cos(\arcsin 0.5 )} }\)

Knowing the Pythegorean identity,
\(\color{#000000 }{ \displaystyle \sin^2\theta +\cos^2\theta =1 }\)
we can obtain that
\(\color{#000000 }{ \displaystyle \cos^2\theta =1 - \sin^2\theta }\)
\(\color{#000000 }{ \displaystyle \cos \theta =\sqrt{1 - \sin^2\theta} }\)

And we can apply the above to our denominator

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{1 - \sin^2(\arcsin 0.5 )}} }\)

Same as by the cosine,
\(\cos^n(\arccos x)=x^n\)

you have the same by the sine function,
\(\sin (\arcsin x)=x \) as well as \(\sin^n(\arcsin x)=x^n\)
(same as by the cosine) ... And we get,

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{1 - \sin^2(\arcsin 0.5 )}} }\)

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{1 - (0.5)^2}} }\)

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{1 - 0.25}} }\)

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{ 0.75}} }\)

and then the rest is a matter of algebra,

\(\color{#000000 }{ \displaystyle \frac{0.5}{\sqrt{ 75/100}} = \frac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=1/\sqrt{3} =\color{blue}{\sqrt{3}/3} }\)

Answer 6

\(\color{#000000 }{ \displaystyle \sin(\arctan e) }\)

We need to express sine in terms of tangent,

\(\tan \theta=\sin\theta/\cos \theta \)
\(\sin\theta=\tan \theta\cos \theta \)
\(\sin\theta=\tan \theta\left(\dfrac{1}{\sec \theta }\right) \)

\(\color{royalblue}{\sec^2\theta=\tan^2\theta+1}\)
\(\color{royalblue}{\sec\theta=\sqrt{\tan^2\theta+1}}\)

\(\sin\theta=\dfrac{\tan \theta}{\sqrt{1-\tan^2\theta } } \)

therefore,

\(\color{#000000 }{ \displaystyle \sin(\arctan e)=\dfrac{\tan (\arctan e)}{\sqrt{1-\tan^2(\arctan e)~ } {\huge\color{white}{|}}} }\)

the rules of tangent are same by cosine and sine
(and by every trig function)
That is, that: \(\tan(\arctan x)=x \),
\(\tan^2(\arctan x)=x^2 \) and any, \(\tan^n(\arctan x)=x^n \)

Therefore, we get,
\(\color{#000000 }{\sin(\arctan e)=\dfrac{e}{\sqrt{1-e^2 }} }\)