39this one is a well known q , why don u give it a try yourself?
orelse post till where u r able to do so that others can continue where u have stopped.
in that way u would learn more
8 Answers
39
13573sinθ-4cosθ=5 and 3cosθ+4sinθ=K
or (3/5)sinθ-(4/5)cosθ=1 and (3/5)cosθ+(4/5)sinθ=K/5
let (3/5)=cosφ, then (4/5)=sinφ
so the expression becomes
sinθcosφ-sinφcosθ=1 and cosθcosφ+sinφsinθ=K/5
or sin(θ-φ)=1=sin90 and cos(θ-φ)=K/5)
But sin(θ-φ)=1 implies cos(θ-φ)=0
Hence, cos(θ-φ)=K/5)=0 and so K=0
1Another way........i don't know if itz right or nt
3sinθ-4cosθ=5
Taking derivative
3cosθ+4sinθ=0
13sinθ-4cosθ=5 --------(1)
3cosθ+4sinθ=K--------(2/
square and add
25=K2 +25
K=0