If ${cos}^{2} 20^{\circ} - {sin}^{2} 20^{\circ} = p$ $cos^(2)20^@-sin^(2)20^@=p$ then ${sin 80}^{\circ} =$ $sin80^@=$ ?

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If ${cos}^{2} 20^{\circ} - {sin}^{2} 20^{\circ} = p$ $cos^(2)20^@-sin^(2)20^@=p$ then ${sin 80}^{\circ} =$ $sin80^@=$ ?

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Answer 1 · 2017-12-02T09:23:00

${sin 80}^{\circ} = \frac{1}{2} \sqrt{2 + \sqrt{2 + 2 p}}$ $sin80^@=1/2sqrt(2+sqrt(2+2p))$

Explanation:

As $cos 2 A = {cos}^{2} A - {sin}^{2} A = 2 {cos}^{2} A - 1 = 1 - 2 {sin}^{2} A$ $cos2A=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A$

${cos}^{2} 20^{\circ} - {sin}^{2} 20^{\circ} = cos (2 \times 20^{\circ}) = {cos 40}^{\circ} = p$ $cos^2 20^@-sin^2 20^@=cos(2xx20^@)=cos40^@=p$

Also $cos A = \sqrt{\frac{1 + cos 2 A}{2}}$ $cosA=sqrt((1+cos2A)/2)$

Hence $2 {cos}^{2} 20^{\circ} - 1 = p$ $2cos^2 20^@-1=p$ and ${cos 20}^{\circ} = \sqrt{\frac{1 + p}{2}}$ $cos20^@=sqrt((1+p)/2)$

and ${cos 10}^{\circ} = \sqrt{\frac{1 + {cos 20}^{\circ}}{2}} = \sqrt{\frac{1 + \sqrt{\frac{1 + p}{2}}}{2}}$ $cos10^@=sqrt((1+cos20^@)/2)=sqrt((1+sqrt((1+p)/2))/2)$

Now ${sin 80}^{\circ} = cos (90^{\circ} - 10^{\circ}) = {cos 10}^{\circ}$ $sin80^@=cos(90^@-10^@)=cos10^@$

= $\sqrt{\frac{1 + \sqrt{\frac{1 + p}{2}}}{2}}$ $sqrt((1+sqrt((1+p)/2))/2)$

= $\sqrt{\frac{\sqrt{2} + \sqrt{1 + p}}{2 \sqrt{2}}}$ $sqrt((sqrt2+sqrt(1+p))/(2sqrt2))$

= $\frac{1}{2} \sqrt{2 + \sqrt{2 + 2 p}}$ $1/2sqrt(2+sqrt(2+2p))$

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If ${cos}^{2} 20^{\circ} - {sin}^{2} 20^{\circ} = p$ $cos^(2)20^@-sin^(2)20^@=p$ then ${sin 80}^{\circ} =$ $sin80^@=$ ?

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Explanation:

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