Math Editor

Further Mathematics

prove that

cos3x+sin3x/cosx+sinx

[(cosx+sinx)(cos2x−cosxsinx+sin2x)]

cosx+sinx

[(cos2x+sin2x−cosxsinx)(cosx+sinx)]

cosx+sinx

1−cosxsinx

Q E D

#2 . If asinθ+b cosθ=c

prove that acosθ−bsinθ=√(a2+b2+c2)​

solution

(acosθ+bsinθ)2=c2

a2sin2θ+b2sin2θ+2absinθcosθ=c2

a2(1−cos2θ)+b2(1−sin2θ)+2absinθcosθ=c2

a2−a2cos2θ+b2−b2sin2θ+2absinθcosθ=c2

a2+b2−c2=acos2θ+bsin2θ−2absinθcosθ

√a2+b2−c2​=(acosθ−bsinθ). Q.E.D

(secA sinA)(secA−cosA) = 1/tanAcot A