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Question

sin47°cos43°2 + cos43° sin47°2 -4cos2 45°

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Solution

sin 47°cos 43°2+cos 43°sin 47°2-4cos245°= sin (90-43)°cos 43°2+ cos 43°sin (90-43)°2-4122 (From the table)= cos 43°cos 43°2+ cos 43°cos 43°2-412 [sin (90-θ)=cos θ]= 12+12-2 = 0

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