Question

what is the cos4x-sin4x​

Answer 1

Answer:

To prove that  

cos

4

x

−

sin

4

x

=

1

−

2

sin

2

x

, we'll need the Pythagorean identity and a variation on the Pythagorean identity:

⇒

cos

2

x

+

sin

2

x

=

1

⇒

cos

2

x

=

1

−

sin

2

x

Step-by-step explanation:

L

H

S

=

cos

4

x

−

sin

4

x

 

L

H

S

=

(

cos

2

x

)

2

−

(

sin

2

x

)

2

L

H

S

=

(

cos

2

x

+

sin

2

x

)

(

cos

2

x

−

sin

2

x

)

L

H

S

=

1

⋅

(

cos

2

x

−

sin

2

x

)

L

H

S

=

cos

2

x

−

sin

2

x

L

H

S

=

1

−

sin

2

x

−

sin

2

x

L

H

S

=

1

−

2

sin

2

x

L

H

S

=

R

H

S

That's the proof. Hope this helped!

Answer 2

answer is given below

we can use here the formula between sin and cos

♒♒♒♒♒♒♒♒♒

$we \: know \: that \\ sin \alpha = sin(90 - \beta ) \\ = \cos( \beta ) \\ so \: here \: we \: can \: compair \: 4x \: as \: \alpha \\ then \\ from \: the \: question \\ cos4x - sin4x \\ = \cos(90 - 4x) - sin \: 4x \\ = \sin(4x) - \sin(4x) \\ = 0 \\ if \: here \: change \: \sin(4x) \\ then \\ cos4x - sin4x \\ = cos4x - sin(90 - 4x) \\ = cos4x - cos4x \\ = 0$

♒♒♒♒♒♒♒♒♒

in the other hand

we can change in other tactic

♒♒♒♒♒♒♒♒♒

$cos4x - sin4x \\ = \frac{1}{sec4x} - sin4x \\ = \frac{1 - sin4x \times sec4x}{sec4x} \\ = \frac{1 - sin4x \times \frac{1}{cos4x} }{sec4x} \\ = \frac{1 - sin4x \times \frac{1}{cos(90 - 4x)} }{sec4x} \\ = \frac{1 - sin4x \times \frac{1}{sin4x} }{sec4x} \\ = \frac{1 - 1}{sec4x} \\ = 0$

♒♒♒♒♒♒♒♒♒

the answer is 0 ✔✔✔✔✔

✨✨hope it helps✨✨