Question

If sin(x-a) = ksin(x+a), express tanx in terms of k

Answer 1

note \[\sin(x-a) = sin(x)cos(a)-cos(x)sin(a)\]
\[\sin(x+a)=sin(x)cos(a)+cos(x)sin(a)\]
then we have \[\sin(x)\cos(a)-\cos(x)\sin(a)=k(\sin(x)\cos(a)+cos(x)\sin(a))\]

distribute the k so we get
\[\sin(x)\cos(a)-\cos(x)\sin(a)=ksin(x)\cos(a)+kcos(x)\sin(a)\] can you finish it off?

Answer 2

Hint: Factoring

Answer 3

no

Answer 4

i don't get it!

Answer 5

If you factor both sides of the last line you get \[\sin(x)\cos(a)-ksin(x)\cos(a)=kcos(x)\sin(a)+\cos(x)\sin(a)\] now factor both sides \[\sin(x)\cos(a)(1-k)=\cos(x)\sin(a)(1+k)\]\[tanx = \left( \frac{ 1+k }{ 1-k } \right)\tan(a)\]

Answer 6

Correction `If you factor both sides of the last line you get` should say if you collect `like terms` of both sides of the last line