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# Writing this expression as a single trig function?

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The expression: 8sin($${\pi \over 3}$$)cos($${\pi \over 3}$$)

I used the double angle identity  sin2A = 2sinAcosA but I'm not sure how I could get the textbook answer of 4sin($${2 \pi \over 3}$$)?

Feb 20, 2019

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8 sin pi/3  cos pi/3           a = pi/3

4  * 2  sin pi/3  cos pi/3                    2 a will equal 2pi/3

In your identity      Sin2a = 2 sin a cos a       but your example is multiplied by 4

4 sin 2a = 4 * 2 sina cos a        See where the 4 comes from now?

4 sin (2pi/3)

Feb 20, 2019