Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15. I let tan15=p, tan30=(2p)/(1-p^2) 1/(√3)= 2p/(1-p^2) 1-p^2= 2√3p p^2 + 2√3p-1=0 I still can't get the answer..
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Originally Posted by Trefoil2727 Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15. I assume that then .
Now if we have , so And since that means , which means the solution is .
Tan x = p, sin 2x = 2p/(1+p^2) let x = 15, Then we have sin 30 = 2p /(1+p^2) = ½ that gives p^2-4p+1=0 Thus p = (4 ±√(16-12))/2 = (4 ±2√3)/2= 2 ±√3 Because 0 < 15 < 90 that is in first quadrant, we have tan 15 = = 2- √3
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