Question

decompose into real factors___trigonometry

Answer 1

\[I=\large 1-\sin^5x-\cos^5x\]

Answer 2

attempt \[1=\sin^2x+\cos^2x-\sin^5x-\cos^5x=\sin^2x(1-\sin^3x)+\cos^2x(1-\cos^3x)\]

Answer 3

typo not 1 but \[I\] on the left

Answer 4

But haven't you already done it?

Answer 5

this is not factorised fully,its still in sum form

Answer 6

\[I=(1-\cos^2x)(1-\sin^3x)+\cos^2x(1-\cos^3x)\]
\[I=1-\sin^3x+\cos^2x(\sin^3x-\cos^3x)\]
now i use \[x^3-y^3=(x-y)(x^2+xy+y^2)\]

Answer 7

\[I=(1-\sin x)(1+\sin x+\sin^2x)+\cos^2x(\sin^2x+\sin x\cos x+\cos^2x)\]
\[I=(1-\sin x)(1+\sin x+\sin^2x)+\cos^2x(1+\sin x\cos x)\]

Answer 8

\[I=(1-\sin x)\left((1-\sin x)(1+\sin x+\sin^2x)+(1+\sin x)(1+\sin x\cos x))\right )\]

Answer 9

i think this is it!!@