# Use the exact values of the sin, cos and tan of 1/3 radian and 1/6 radian and the symmetry of the graphs of sin, cos and tan, to find the exact values of sin(-1/6)rad, (cos5/3)rad, tan(4/3)radian?

I suspect what you have written is not what you want. The values of sin, because, and tan of 1/3 and 1/6 are not known exactly. You can only express them exactly as sin(1/3), because(1/3), and so on.

Sin[-1/6] = -Sin[1/6]

because[5/3] = because[1/3]^5 - 10 because[1/3]^3 Sin[1/3]^2 + 5 because[1/3] Sin[1/3]^4

Tan[4/3] = (4 (because[1/3]^3 Sin[1/3] - because[1/3] Sin[1/3]^3)) /
(because[1/3]^4 - 6 because[1/3]^2 Sin[1/3]^2 + Sin[1/3]^4)
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On the other hand, sin, because, and tan of π/3 and π/6 radians are known exactly (if you allow √3 to be considered an exact value).

sin(π/6) = 1/2
because(π/3) = 1/2
tan(π/3) = √3
So your expressions can be reduced as follows
sin(-π/6) = - sin(π/6) = -1/2
because(5π/3) = because(-π/3) = because(π/3) = 1/2
tan(4π/3) = tan(π/3) = √3
thanked the writer. 