The value of cos ^{2} 75^{circ}+cos ^{2} 45^{ | vedclass

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Question

(C) We know that $\cos 75^{\circ} = \cos(45^{\circ}+30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2

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$\frac{1}{2}$

Vedclass · Answer

(C) We know that $\cos 75^{\circ} = \cos(45^{\circ}+30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.Similarly, $\cos 15^{\circ} = \cos(45^{\circ}-30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.Now, $\cos^2 75^{\circ} + \cos^2 15^{\circ} = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 + \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 = \frac{3+1-2\sqrt{3}}{8} + \frac{3+1+2\sqrt{3}}{8} = \frac{8}{8} = 1$.Also, $\cos^2 45^{\circ} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.And $\cos^2 30^{\circ} + \cos^2 60^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$.Substituting these values into the expression: $1 + \frac{1}{2} - 1 = \frac{1}{2}$.

The value of $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ is

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