Trigonometry Homework

Question

Solve this equation for exact solutions over the interval [ 0, 2𝜋 ) :

2 cos 𝜃 sin 𝜃 = 0

Answer 1

Given:

2 sin θ cos θ = 0

θ : [ 0, 2π )

Find:

θ = ? rad

Solution:

From either a Google search or from class notes or the like, a well-known trig identity can help transform this problem into something simpler.

sin 2θ = 2 sin θ cos θ

Thus,

sin 2θ = 0

Taking the inverse sine on both sides results in:

2θ = sin^-1( 0 ) = πn n ∈ (this is fancy math lingo that n can be ANY real integer, like 1, 2, 3, 4, or so on)

θ = πn / 2 = 0, π/2, π, 3π/2, ...

However, θ is being limited to no less than or equal to 0 and no greater than 2π

Thus,

θ = 0, π/2, π, 3π/2

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

As a good gut check, plug these numbers back into the original problem:

2 sin( 0 ) cos( 0 ) = 2 ( 0 ) ( 1 ) = 0 ≡ 0 (again, this ≡ is fancy math lingo that the left side of the equation that has been calculated to be zero is defined to be equal to the right side of the equation, which is also zero, what we wanted from the start!)

2 sin( π/2 ) cos( π/2 ) = 2 ( 1 ) ( 0 ) = 0 ≡ 0

2 sin( π ) cos( π ) = 2 ( 0 ) ( 1 ) = 0 ≡ 0

2 sin( 3π/2 ) cos( 3π/2 ) = 2 ( 1 ) ( 0 ) = 0 ≡ 0

Answer 2

2cosTsinT=0

use trig identity: 2cosTsinT = sin2T

sin2T = 0

2T = sin^-1(0)

= 0 or pi or 2pi or 4pi

Theta = 0/2 or pi/2

=0, pi/2, 3pi/2, or pi

check the answers

2cosTsinT=0

divide by 2

cosTsinT = 0

cos(0)sin(0) =1(0) = 0

cos(pi/2)sin(pi/2) = 0(1) = 0

2cos(pi)(sin(pi)) = 2(1)(0) = 0

2cos(3pi/2)(sin(3pi/2)) = 2(0)(-1) = 0