Math, asked by krish722687, 1 year ago

what is the cos4x-sin4x​

Answers

Answered by omsdpandey
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!


william22: you are dum
Answered by attinderpaul55225
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✨✨

Similar questions