Using $ an 2 heta = 2 an heta over 1 - an ^2 heta $ v $ heta = pi over 16$:

$eqalign & an pi over 8 = 2 an pi over 16 over 1 - an ^2pi over 16 cr & an pi over 8 = 0.41421.... cr $

I basically get solution equivalent come $sqrt 2 - 1$ but not an "exact" answer, exactly how do i go about this inquiry to get a result in "exact" form?

Thank you.

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algebra-precalculus trigonometry

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request Apr 4 "13 in ~ 23:26

seekerseeker

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## 4 answers 4

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We have $ an(2 heta) = dfrac2 an( heta)1- an^2( heta)$. Take it $ heta = dfracpi8$ and also let $t = an(pi/8)$. We then get$$ an(pi/4) = dfrac2 an(pi/8)1- an^2(pi/8) = dfrac2t1-t^2$$Hence, us get$$1-t^2 = 2t implies (t+1)^2 = 2 implies t = -1 pm sqrt2$$Since $t > 0$, we get that $ an(pi/8) = sqrt2-1$

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answer Apr 4 "13 in ~ 23:30

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You"re going about this the wrong way; instead, collection $ heta = fracpi8$. You get:

$$ an fracpi4 = 1 = frac2 an fracpi81- an^2 fracpi8$$

Multiply through by $1- an^2 fracpi8$ and solve the result quadratic equation.

If the helps simplify notation, substitute $t= an^2 fracpi8$. The equation has actually two roots, one positive and also one negative, therefore you"ll require to give reasoning about which come pick.

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answer Apr 4 "13 at 23:30

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An identification that is useful when considering the stereographic projection and also the half-tangent integration substitution:$$ an(x/2)=fracsin(x)1+cos(x)$$We know that $sin(pi/4)=cos(pi/4)=1/sqrt2$. Therefore,$$eginalign an(pi/8)&=frac1/sqrt21+1/sqrt2frac1-1/sqrt21-1/sqrt2\&=sqrt2-1endalign$$

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reply Apr 4 "13 at 23:48

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$$sqrt2-1=cscdfracpi4-cotdfracpi4=dfrac1-cosdfracpi4sindfracpi4=dfrac2sin^2dfracpi82sindfracpi8cosdfracpi8=?$$ making use of $sin2A=2sin Acos A,cos2B=1-2sin^2B$

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reply Oct 23 "19 at 7:12

laboratory bhattacharjeelab bhattacharjee

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