Question

√(sinx-√(sinx+cosx))=cosx

Answer 1

what is the actual question? they are not equal

Answer 2

These are equal only when

$$
x = \cfrac{1}{2} (4 \pi n+\pi), n \in \mathbb{Z}
$$

Answer 3

you must prove that they are equal

Answer 4

here is my proof: let \(x=0\) then the left hand side is \(-1\) and the right hand side is \(1\) so they are definitely not equal

Answer 5

can you write it step by step?

Answer 6

$$
\sin(x)-\sqrt{\sin x + \cos x}=\cos^2x\\
-\sqrt{\sin x + \cos x}=\cos^2x-\sin x\\
\sin x + \cos x=\left (\cos^2x-\sin x\right )^2\\
\sin x + \cos x=\cos^4x-2\cos^2x\sin x+\sin^2x\\
1=\cfrac{\cos^2x(\cos^2x-2\sin x)+\sin^2x}{\sin x + \cos x}\\
\implies 1=\cfrac{\cos^2x-2\sin x}{\sin x+\cos x}\text{ and }\\
\qquad 1=\cfrac{1}{\sin x+\cos x}
$$
Which can only occur if
$$
x = \cfrac{1}{2} (4 \pi n+\pi), n\in \mathbb{Z}.
$$

Answer 7

Yes, (got disconected) what satelline said, but differently
let x=90
Cos90=0
Sin90=1
So we get 1=0

Answer 8

\[you wrote 1=(\cos^2⁡x (\cos^2 x-2sinx)+\sin^2⁡x)/(sinx+cosx) then 1=\frac{ \cos^2⁡x-\sin^2⁡x}{ sinx+cosx} how?\]

Answer 9

you can use ~ as a space in equations.
Also, try to substitute 90 for x, that's what satellite was talking about.

( Sin90=1 and Cos90=0 )

Answer 10

no you can't to substitute 90 for x,you must to prove this

Answer 11

you should solve this

Answer 12

maybe I was reading the problem wrong?\[\sqrt{Sinx-\sqrt{Sinx+Cosx}}=Cosx\]

Answer 13

this?

Answer 14

yes it's right

Answer 15

square both sides\[sinx-\sqrt{sinxcosx}=\cos^2x\]
subtract sinx from both sides\[-\sqrt{sinxcosx}=Cos^2x-Sinx\]
\[So~~far~~so~~good?\]

Answer 16

yes

Answer 17

\[Squar e~~both~~sides~~~~~~~~sinxcosx=Cos^4x-2Cos^2xSinx+Sin^2x\]

Answer 18

where is +? sinx+cosx

Answer 19

- is also squared.

Answer 20

(-2)^2=4 not -4

Answer 21

Right?

Answer 22

yes, but you right sinxcosx it' false you didn't write + sinx+cosx

Answer 23

When there is a number, let say 2 it is same as +2.

Answer 24

\[\sin x+\cos x\]=\[\cos^{4} x-2\cos^{2} x \sin +\sin^{2} \] then?

Answer 25

that's the problem?

Answer 26

I simplified till\[sinxcosx=\cos^4x+\cos^2xsinx+Sin^2x\]

Answer 27

ok

Answer 28

\[-sinx^2+sinxcosx=\cos^4x+\cos^2xsinx\]\[-sinx^2+sinxcosx~~~~~~~~->~~~~~~~~-sinx(sinx-cosx)\]divide both sides by sinx\[-sinx+\cos=\cos^4/sinx~~+~~\cos^2x\]\[-sinx+cosx=\cos^3xcotx+\cos^2x\]

Answer 29

\[-sinx=\cos^3cotx+\cos^2x-cosx\]\[-sinx=cosx(\cos^2xcotx+cosx-1)\]\[-tanx=\cos^2xcotx+cosx-1\]

Answer 30

\[-tanx=cosx(cotx+cosx-secx)\]\[-sinx=cotx+cosx-secx\]

Answer 31

it's kind of taking too long, ik.

Answer 32

\[\sqrt{\sin x-\sqrt{sinx+\cos x }}=cosx\]

Answer 33

it's prove?

Answer 34

no.you'll solve this