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Mathematics Question on Trigonometric Equations

cos⁡25∘−cos⁡215∘−sin⁡215∘+sin⁡235∘\cos ^{2} 5^{\circ}-\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}+\sin ^{2} 35^{\circ} +cos⁡15∘sin⁡15∘−cos⁡5∘sin⁡35∘=+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ}=

A

0

B

1

C

32\frac{3}{2}

D

2

Answer

0

Explanation

Solution

cos⁡25∘−cos⁡215∘−sin⁡215∘+sin⁡235∘\cos ^{2} 5^{\circ}-\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}+\sin ^{2} 35^{\circ}
+cos⁡15∘sin⁡15∘−cos⁡5∘sin⁡35∘+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ}
=cos⁡5∘(cos⁡5∘−sin⁡35∘)−cos⁡15∘(cos⁡15∘−sin⁡15∘)=\cos 5^{\circ}\left(\cos 5^{\circ}-\sin 35^{\circ}\right)-\cos 15^{\circ}\left(\cos 15^{\circ}-\sin 15^{\circ}\right)
+sin⁡235∘−sin⁡215∘+\sin ^{2} 35^{\circ}-\sin ^{2} 15^{\circ}
=cos⁡5∘(cos⁡5∘−cos⁡55∘)−cos⁡15∘(cos⁡15∘−cos⁡75∘)=\cos 5^{\circ}\left(\cos 5^{\circ}-\cos 55^{\circ}\right)-\cos 15^{\circ}\left(\cos 15^{\circ}-\cos 75^{\circ}\right)
+sin⁡50∘sin⁡20∘+\sin 50^{\circ} \sin 20^{\circ}
=cos⁡5∘(2sin⁡30∘sin⁡25∘)−cos⁡15∘(2sin⁡45∘sin⁡30∘)=\cos 5^{\circ}\left(2 \sin 30^{\circ} \sin 25^{\circ}\right)-\cos 15^{\circ}\left(2 \sin 45^{\circ} \sin 30^{\circ}\right)
+sin⁡50∘sin⁡20∘+\sin 50^{\circ} \sin 20^{\circ}
=−cos⁡15∘2+cos⁡5∘sin⁡25∘+sin⁡50∘sin⁡20∘=-\frac{\cos 15^{\circ}}{\sqrt{2}}+\cos 5^{\circ} \sin 25^{\circ}+\sin 50^{\circ} \sin 20^{\circ}
=−cos⁡15∘2+12(2cos⁡5∘cos⁡65∘)+12(2sin⁡50∘sin⁡20∘)=-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{1}{2}\left(2 \cos 5^{\circ} \cos 65^{\circ}\right)+\frac{1}{2}\left(2 \sin 50^{\circ} \sin 20^{\circ}\right)
=−cos⁡15∘2+12[cos⁡70∘+cos⁡60∘+cos⁡30∘−cos⁡70∘]=-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{1}{2}\left[\cos 70^{\circ}+\cos 60^{\circ}+\cos 30^{\circ}-\cos 70^{\circ}\right]
=−cos⁡5∘2+12[2cos⁡45∘cos⁡15∘]=-\frac{\cos 5^{\circ}}{\sqrt{2}}+\frac{1}{2}\left[2 \cos 45^{\circ} \cos 15^{\circ}\right]
=−cos⁡15∘2+cos⁡15∘2=0=-\frac{\cos 15^{\circ}}{\sqrt{2}}+\frac{\cos 15^{\circ}}{\sqrt{2}}=0