Exercise 32 Page 963

We want to solve the following equation. sin ( θ - π/2 ) = sec θ We will first simplify the left-hand side of the situation. To do so, we need to consider which Trigonometric Identities will be useful in this equation. Let's recall the Angle Difference Identity for the sine. sin ( A - B ) = sin A cos B - cos A sin BWith this relationship in mind, let's simplify the left-hand side of the equation.

sin ( θ - π/2 )

sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

sin θ cos π/2 - cos θ sin π/2

Distr

Distribute -1

sin θ cos π/2 - cos θ sin π/2

cos π/2 = 0

sin θ (0) - cos θ sin π/2

ZeroPropMult

Zero Property of Multiplication

-cos θ sin π/2

IdPropAdd

Identity Property of Addition

-cos θ sin π/2

sin π/2 = 1

-cos θ (1)

IdPropMult

Identity Property of Multiplication

-cos θ

We have proven the following identity. sin ( θ - π/2 ) = -cos θ Let's use it to simplify the left-hand side of our equation. sin ( θ - π/2 ) &= sec θ &⇕ -cos θ &= sec θ Notice that we now have a much simpler equation. Now, let's recall the Secant Identity sec θ = 1/cos θ Let's use this identity to express our equation solely in terms of cos θ. However, note that since we have a cos θ in the denominator, we need to check whether for any solutions we might find cos θ is non-zero.

- cos θ = sec θ

sec(θ) = 1/cos(θ)

- cos θ = 1/cos θ

MultEqn

LHS * cos θ=RHS* cos θ

- cos^2 θ = 1

MultEqn

LHS * - 1 =RHS* - 1

cos^2 θ = - 1

Now, notice that the left-hand side contains a square, while the right-hand side is negative. This means that our equation has no solution, as no real number multiplied by itself becomes negative.