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5 January, 02:37

Sin 70°cos 10°-cos 70°sin 10=√3/2

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  1. 5 January, 04:59
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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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