Asked by Ritu
Cos3Q=√3/2,find tan6Q?
Answers
Answered by
Steve
if cos3Q = √3/2,
sin3Q = ±1/2
so, tan3Q = ±1/√3
so, using the +1/2 value,
tan6Q = 2tan3Q/(1-tan^2 3Q)
= 2(1/√3)/(1 - 1/3)
= (2/√3)/(2/3)
= 3/√3
= √3
so, -√3 if sin3Q = -1/2
sin3Q = ±1/2
so, tan3Q = ±1/√3
so, using the +1/2 value,
tan6Q = 2tan3Q/(1-tan^2 3Q)
= 2(1/√3)/(1 - 1/3)
= (2/√3)/(2/3)
= 3/√3
= √3
so, -√3 if sin3Q = -1/2
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