Question

50 points need help asap plzz!!! Show that the Pythagorean identity sin^2 θ+cos^2 θ=1 is true for the given angle.

θ= 5π/3

Answer 1

Pythagorean identity sin²θ+cos²θ = 1 is true for the angle θ =
`(5\pi )/(3)`

Explanation:

At first, let us simplify the left side of the identity

∵ The left side is sin²Ф + cos²Ф

∵ Ф =
`(5\pi )/(3)` ⇒ lies in the 4th quadrant

∴ The left side is sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`)

→ Let us write the values of sin(
`(5\pi )/(3)`) and cos(
`(5\pi )/(3)`)

∵ sin(
`(5\pi )/(3)`) =
`(-√(3))/(2)` ⇒ sine an angle in the 4th quadrant is -ve

∵ cos(
`(5\pi )/(3)`) =
`(1)/(2)` ⇒ cosine an angle in the 4th quadrant is +ve

→ Substitute them in the left side

∵ sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`) = [
`(-√(3))/(2)`]² + [
`(1)/(2)`]²

∴ sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`) = [
`(3)/(4)`] + [
`(1)/(4)`]

∴ sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`) = [
`(4)/(4)`]

∴ sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`) = 1

∵ The right side = 1

∴ Left side = Right side

∴ sin²(
`(5\pi )/(3)`) + cos²(
`(5\pi )/(3)`) = 1 ⇒ proved

∴ Pythagorean identity sin²θ+cos²θ = 1 is true for the angle θ =
`(5\pi )/(3)`