The value of cos^2 10^o - cos 10^o cos 50^o + | vedclass

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(B) Let $E = \cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$. Multiplying by $\frac{2}{2}$,we get $E = \frac{1}{2} (2 \cos^2 10^o - 2 \cos 10^o \cos

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$\frac{3}{4}$

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(B) Let $E = \cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$. Multiplying by $\frac{2}{2}$,we get $E = \frac{1}{2} (2 \cos^2 10^o - 2 \cos 10^o \cos 50^o + 2 \cos^2 50^o)$. Using the identities $2 \cos^2 	heta = 1 + \cos 2	heta$ and $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $E = \frac{1}{2} [(1 + \cos 20^o) - (\cos 60^o + \cos(-40^o)) + (1 + \cos 100^o)]$. $E = \frac{1}{2} [2 + \cos 20^o - \frac{1}{2} - \cos 40^o + \cos 100^o]$. $E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \sin(\frac{100^o+40^o}{2}) \sin(\frac{100^o-40^o}{2})]$. $E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \sin 70^o \sin 30^o]$. Since $\sin 30^o = \frac{1}{2}$ and $\sin 70^o = \cos 20^o$: $E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - 2 \cos 20^o \cdot \frac{1}{2}]$. $E = \frac{1}{2} [\frac{3}{2} + \cos 20^o - \cos 20^o] = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$.

The value of $\cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$ is

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