Sin (x-20)° =cos (5x-10)°​

sin(x - 20)° = cos(5x - 10)°

Expand 5x - 10 in terms of x - 20:

5x - 10 = 5x - 100 + 90

… = 5 (x - 20) + 90

and recall that for all θ, cos(θ + 90)° = - sin(θ)°, so

sin(x - 20)° = cos(5 (x - 20) + 90)°

sin(x - 20)° = - sin(5 (x - 20))°

Replace (x - 20)° = y ° to make this a bit easier to read:

sin(y °) = - sin(5y )°

Expand the right side in terms of powers of sin:

sin(5θ) = 5 sin(θ) - 20 sin³(θ) + 16 sin⁵(θ)

so the equation becomes

sin(y °) = -5 sin(y °) + 20 sin³(y °) - 16 sin⁵(y °)

16 sin⁵(y °) - 20 sin³(y °) + 6 sin(y °) = 0

Replace z = sin(y °) to get a polynomial equation:

16z ⁵ - 20z ³ + 6z = 0

Factorize the left side:

2z (8z ⁴ - 10z ² + 3) = 0

2z (2z ² - 1) (4z ² - 3) = 0

Solve for z :

• 2z = 0   →   z = 0

• 2z ² - 1 = 0   →   z ² = 1/2   →   z = ± 1/√2

• 4z ² - 3 = 0   →   z ² = 3/4   →   z = ± √3/2

Solve for y in each case above:

• z = sin(y °) = 0   →   y = 0 + 360n   or   y = 180 + 360n

(where n is any integer)

These solutions can be combined into one family, y = 180n.

• z = sin(y °) = -1/√2   →   y = 225 + 360n   or   y = 315 + 360n

• z = sin(y °) = 1/√2   →   y = 45 + 360n   or   y = 135 + 360n

These solutions can also be combined, y = 45 + 90n.

• z = sin(y °) = -√3/2   →   y = 240 + 360n   or   y = 300 + 360n

• z = sin(y °) = √3/2   →   y = 60 + 360n   or   y = 120 + 360n

These can be combined as y = 60 + 180n or y = 120 + 180n.

Solve for x in each case:

• y = x - 20 = 180n   →   x = 20 + 180n

• y = x - 20 = 45 + 90n   →   x = 65 + 90n

• y = x - 20 = 60 + 180n   →   x = 80 + 180n

• y = x - 20 = 120 + 180n   →   x = 140 + 180n

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

addition

step-by-step explanation: