Question

if a cos theta- b sin theta =c show that +- root of a square + b square - c square

Answer 1

I am considering theta as 't' for the purpose of clarity...

First square both the sides of provided equation.
==> {acost - bsint}^2 = c^2
==> a^2(cost)^2 + b^2(sint)^2 -2abcostsint=c^2---eq(1)
Now consider LHS of the equation which should be proved right.Let that be equal to 't'

y= {asint+bcost}
y^2=a^2(sint)^2 + b^2(cost)^2 + 2absintcost---eq(2)

By adding equations eq(1) and eq(2) we get

a^2{(sint)^2 + (cost)^2} + b^2{ (sint)^2 + (cost)^2}=y^2 + c^2

==> a^2 + b^2 = y^2 + c^2
===>y=+/-
$\sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$
Hence proved