cos ^2 5^{circ}-cos ^2 15^{circ}-sin ^2 15^{c | vedclass

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Question

(A) Let the expression be $E = \cos ^2 5^{\circ}-(\cos ^2 15^{\circ}+\sin ^2 15^{\circ})+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\c

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let the expression be $E = \cos ^2 5^{\circ}-(\cos ^2 15^{\circ}+\sin ^2 15^{\circ})+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ}$.
Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have $\cos^2 15^{\circ} + \sin^2 15^{\circ} = 1$.
Also,using $2 \sin 	heta \cos 	heta = \sin 2	heta$,we have $\cos 15^{\circ} \sin 15^{\circ} = \frac{1}{2} \sin 30^{\circ} = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.
Substituting these into $E$,we get $E = \cos^2 5^{\circ} - 1 + \sin^2 35^{\circ} + \frac{1}{4} - \cos 5^{\circ} \sin 35^{\circ}$.
Since $\cos^2 5^{\circ} - 1 = -\sin^2 5^{\circ}$,we have $E = -\sin^2 5^{\circ} + \sin^2 35^{\circ} + \frac{1}{4} - \cos 5^{\circ} \sin 35^{\circ}$.
Using $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B)$,we have $\sin^2 35^{\circ} - \sin^2 5^{\circ} = \sin(35^{\circ}-5^{\circ})\sin(35^{\circ}+5^{\circ}) = \sin 30^{\circ} \sin 40^{\circ} = \frac{1}{2} \sin 40^{\circ}$.
Using $2 \cos A \sin B = \sin(A+B) - \sin(A-B)$,we have $\cos 5^{\circ} \sin 35^{\circ} = \frac{1}{2} [\sin(35^{\circ}+5^{\circ}) - \sin(35^{\circ}-5^{\circ})] = \frac{1}{2} (\sin 40^{\circ} - \sin 30^{\circ}) = \frac{1}{2} \sin 40^{\circ} - \frac{1}{4}$.
Substituting these back into $E$,we get $E = \frac{1}{2} \sin 40^{\circ} + \frac{1}{4} - (\frac{1}{2} \sin 40^{\circ} - \frac{1}{4}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} = $

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