Find Sin(2* It has to do with trigonometry, I can't find the formula anywhere, need help asap
1 answer:
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Alright, here it is
w=(q+p)/(q-pq)
factor out the q in the bottom part
w=(q+p)/[(q)(1-p)]
multiply both sides by q
wq=(q+p)/(1-p)
add 1 to both sides, but add (1-p)/(1-p) to the right side since that equals 1
wq+1=(q+1+p-p)/(1-p)=(q+1)/(1-p)
multiply both sdies by (1-p)
(wq+1)(1-p)=q+1
divide both sdies by (wq+1)
1-p=(q+1)/(wq+1)
subtract 1 from both sdies
-p=[(q+1)/(wq+1)]-1
multiply by -1
p=-[(q+1)/(wq+1)]+1 or
w=(q+p)/(q-pq)
factor out the q in the bottom part
w=(q+p)/[(q)(1-p)]
multiply both sides by q
wq=(q+p)/(1-p)
add 1 to both sides, but add (1-p)/(1-p) to the right side since that equals 1
wq+1=(q+1+p-p)/(1-p)=(q+1)/(1-p)
multiply both sdies by (1-p)
(wq+1)(1-p)=q+1
divide both sdies by (wq+1)
1-p=(q+1)/(wq+1)
subtract 1 from both sdies
-p=[(q+1)/(wq+1)]-1
multiply by -1
p=-[(q+1)/(wq+1)]+1 or