Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Angle Identities
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Exercise 16 Page 949

Use the Cofunction Identity cos ( π2-θ)=sin θ.

Ï€/2, 3Ï€/2

Practice makes perfect

Let's start by recalling the Cofunction Identity and the Reciprocal Identity for cosecant. Cofunction Identity cos (π/2-θ)=sin θ [0.8em] Reciprocal Identity for Cosecant csc θ =1/sin θ To solve the given equation, we will first rewrite it using this identity.

cos (π/2-θ )=csc θ
sin θ=csc θ
sin θ = 1/sin θ
sin θ * sin θ = 1/sin θ * sin θ
sin^2 θ = 1/sin θ * sin θ
sin^2 θ = 1
sqrt(sin^2 θ) = sqrt(1)

sqrt(a^2)=± a

sin θ = ± sqrt(1)
sin θ = ± 1
lc sin θ = 1 & (I) sin θ = -1 & (II)

We will solve the obtained equations one at a time.

sin θ=1

We will isolate θ using an inverse trigonometric function. sin θ =1 ⇔ θ =sin ^(- 1) (1) Let's use a calculator to find the value for θ.

θ =sin ^(- 1)1
θ =π/2

Finally, we need to check if there are any other possible angles that satisfy this equation within the given range. Given Range: 0 ≤ θ < 2π Let's consider the unit circle. Recall that the sine of an angle in standard position is the second coordinate of the point of intersection between its terminal side and the circle. Let's plot all the points on the unit circle with second coordinate equal to 1.

As we can see above, the angle whose sine is 1 is π2.

sin θ=-1

We will isolate θ. sin θ =-1 ⇔ θ =sin ^(- 1) (-1) Let's use a calculator to find the value for θ.

θ =sin ^(- 1)(-1)
θ =3π/2

Finally, we need to check if there are any other possible angles that satisfy this equation within the given range. Given Range: 0 ≤ θ < 2π Let's plot all the points on the unit circle with second coordinate equal to -1.

Therefore, the solutions of the given equation in the given range are π2 and 3π2.