On the first, start with the double angle formulas.
On the second, the proof is trivial.
2t-t is t.
sin(2t-t)=sint
prove that [(sin 2 t / sin t )] - [( cos 2t ) / cos t ] = sec t
and sin (2t - t ) = sin t
2 answers
In fact I have solved this like that
follows:
=(sin 2t*cost-cos2t*sint )/cost*sint
= sin (2t-t)/sint*cost
= sint/sint*cost
=1/cost
=sect QED
Where 't' stands for 'theta' .
Was I correct ?
follows:
=(sin 2t*cost-cos2t*sint )/cost*sint
= sin (2t-t)/sint*cost
= sint/sint*cost
=1/cost
=sect QED
Where 't' stands for 'theta' .
Was I correct ?
4 answers