Trigonometric identity

[brinazarski]

Thank you I have a new question... as usual ><

cos²150x-sin²150x = sin5xcos40-cos5xsin40


I have no idea how to solve this... I tried using cos^-1 on both sides and one side was 90, the other, 92.something.

 

[brinazarski]

Can someone help me with my most recent problem? I got that equation from cos(3x + 50) = sin(5x - 40)

 

[Deleted member 4993]

Thank you I have a new question... as usual ><

cos²150x-sin²150x = sin5xcos40-cos5xsin40


I have no idea how to solve this... I tried using cos^-1 on both sides and one side was 90, the other, 92.something.

Hint:

cos²(Θ) - sin²(Θ) = cos(2Θ)

and

sin(Φ)*cos(Θ) - cos(Φ)*sin(Θ) = sin(Φ - Θ)

Please share your work with us - indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Deleted member 4993]

Can someone help me with my most recent problem? I got that equation (cos²150x-sin²150x = sin5xcos40-cos5xsin40) from cos(3x + 50) = sin(5x - 40)

Incorrect

How did you get that?

If you are trying to solve for x in

cos(3x + 50) = sin(5x - 40)

use:

cos(Θ) = sin(π/2 ± Θ)

 

[soroban]

Hello, brinazarski!

$\displaystyle \text{Solve for }x\!:\;\;\cos(3x + 50) \:=\: \sin(5x - 40)$

Identity: .$\displaystyle \cos(\theta) \:=\:\sin(90^o - \theta)$

Hence: .$\displaystyle \cos(3x+50) \:=\:\sin\big[90 - (3x+50)\big] \:=\:\sin(40-3x)$


The equation becomes: .$\displaystyle \sin(40-3x) \:=\:\sin(5x-40)$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle 40-3x \:=\:5x - 40$

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle 8x \:=\:80$

n . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x \:=\:10$

 

[brinazarski]

@ Subhotosh Khan: I got it from the fact that cos(A + B) = cosAcosB - sinAsinB and sin(A - B) = sinAcosB - cosAsinB

cos(3x + 50) = sin(5x - 40)

cos3xcos50 - sin3xsin50 = sin5xcos40 - cos5xsin40

cos²150x - sin²150x = sin5xcos40 - cos5xsin40

That's how I got it :/ Maybe I should've clarified that earlier.

@soroban: "Identity: .

" <- I'm not really sure how to use that identity...

"Hence: .

" <- I don't see how you got sin(40 - 3x) from sin(90-(3x + 50)). Can you explain that please...?

 

[Deleted member 4993]

@ Subhotosh Khan: I got it from the fact that cos(A + B) = cosAcosB - sinAsinB and sin(A - B) = sinAcosB - cosAsinB

cos(3x + 50) = sin(5x - 40)

cos3xcos50 - sin3xsin50 = sin5xcos40 - cos5xsin40

cos²150x - sin²150x = sin5xcos40 - cos5xsin40 <<< How did you get that from above?

That's how I got it :/ Maybe I should've clarified that earlier.

@soroban: "Identity: .

" <- I'm not really sure how to use that identity...

"Hence: .

" <- I don't see how you got sin(40 - 3x) from sin(90-(3x + 50)). Can you explain that please...?

90 - (3x + 50) = 90 - 3x - 50 = 90 - 50 - 3x = 40 - 3x

 

[brinazarski]

cos x cos = cos²

3x x 50 = 150x

sin x sin = sin²

3x x 50 = 150x

"90 - (3x + 50) = 90 - 3x - 50 = 90 - 50 - 3x = 40 - 3x"

^ why did the plus turn into a minus? Otherwise it makes sense, I think

 

[Deleted member 4993]

cos x cos = cos²

3x x 50 = 150x

sin x sin = sin²

3x x 50 = 150x

I was afraid - that is what you meant!!

"90 - (3x + 50) = 90 - 3x - 50 = 90 - 50 - 3x = 40 - 3x"

^ why did the plus turn into a minus? Otherwise it makes sense, I think

You need to review operations with functions.

For starter:

f(x) * f(y) remain f(x) * f(y)

DOES NOT BECOME

f²(xy)

You need serious help from a face-to-face tutor.