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Mathematics
Giselle Macdonald
30 May, 12:03
Sin 70°cos 10°-cos 70°sin 10=√3/2
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Joshua Mercer
30 May, 13:22
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Let a = 70° and b = 10° (and a-b=70-10 = 60)
We have the following trigonometric identity:
sin (a-b) = sin (a). cos (b) - sin (b). cos (a) OR:
sin (70-10) = sin70. cos10 - sin10. cos70
But sin (70-10) = sin (60) and we know that sin (60°) = (√3) / 2
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