show that \$ an pi over 8 = sqrt 2 - 1\$, making use of the identification \$ an 2 heta = 2 an heta over 1 - an ^2 heta \$

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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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
user17762user17762
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\$\$ 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

robjohn♦robjohn
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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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