Question

( cos x - sin x )^ 2 = 1- sin 2 x​

Answer 1

Answer:

It depends on how far back you need to go in your proof.

If you can use:

sin

2

(

x

)

+

cos

2

(

x

)

=

1

(which can be derived from the Pythagorean Theorem)

and

sin

(

2

x

)

=

2

sin

(

x

)

⋅

cos

(

x

)

(the Double Angle Formula for sin, which is considerably more complicate to prove)

then the requested proof is fairly simple:

(

sin

(

x

)

+

cos

(

x

)

)

2

=

sin

2

(

x

)

+

2

sin

(

x

)

⋅

cos

(

x

)

+

cos

2

(

x

)

=

[

sin

2

(

x

)

+

cos

2

(

x

)

]

+

2

sin

(

x

)

⋅

cos

(

x

)

=

1

+

sin

(

2

x

)

Answer 2

$LHS= \: (cos \: x - sin \: x)^{2}$

∴ $a^{2} + {b}^{2} - 2a\times b =(a - b)^{2}$

⇒$cos \: x^{2} + sin \: {x}^{2} - 2cos \: x \times sin \: x$

⇒$sin \: {x}^{2} + cos \: {x}^{2} - 2sinx \times cos \: x$

$1 - sin2x = RHS\: \: \: ∴(sin \: {x}^{2} + cos \: {x}^{2} = 1)$

$\ \: \: \: \: \:∴(2sin \: x \times cos \: x = sin \: 2x)$