Question

sin³+cos³÷sin+cos +sin+cos=1​

Answer 1

sinθ−cosθ=3/4…(1)

(sinθ−cosθ)2=9/16=>sin2θ+cos2θ−2sinθcosθ=9/16=>

1−2sinθcosθ=9/16=> 2sinθcosθ=1−9/16=>

2sinθcosθ=7/16=>sinθcosθ=7/32...(2)

Now, we set x=sinθ, y=cosθ and we take:

sin3θ+cos3θ=x3+y3=(x+y)(x2−xy+y2)...(3)

Moreover, we have:

(x+y)2=(x2+y2+2xy)=>(x+y)2=1+2xy=>

x+y=sqrt(1+2xy)=>x+y=sqrt(1+7/16)...(4)

Therefore, by (3) and (4), we take:

sin3θ+cos3θ=x3+y3=(x+y)(x2−xy+y2) =

sqrt(1+7/16)(1−xy)…(5)

Now, by (2) we take:

xy=7/32...(6)

Finally, by (5) and (6) we obtain:

sin3θ+cos3θ=sqrt(1+7/16)(1−7/32)=

(25/32)sqrt(23/16)=(25/128)sqrt(23)