Simplify this Sine Problem

Question

cos(pi/2 - x)[sec(pi/2 - x)-cos(pi/2 - x)]

Answer 1

Locate and copy of table of frequently used trigonometry conversions.

cos (π/2 - θ)[sec (π/2 - θ) - cos (π/2 - θ)]

sin θ [(1 / (cos (π/2 - θ)) - sin θ]

sin θ [(1 / sin θ) - sin θ]

1 - sin² θ

cos² θ

Answer 2

cos(pi/2 - x)[sec(pi/2 - x)-cos(pi/2 - x)]

Let's say: θ = pi/2 - x

Then: cos(θ)[sec(θ)-cos(θ)]

Now distribute cos(θ) across sec(θ)-cos(θ):

cos(θ)*sec(θ)-cos(θ)*cos(θ)

Using the Reciprocal Identity: sec(θ) = 1/ cos(θ)

cos(θ)*sec(θ)-cos(θ)*cos(θ)

becomes: 1- cos²(θ)

Among the Fundamental Identities is: sin²(θ) +cos²(θ)=1

Thus 1- cos²(θ) = sin²(θ).

Replace θ with pi/2 - x:

sin²(pi/2 - x)

Two options here:

First: sine and cosine are 90 degrees out of phase which corresponds to pi/2;

we can use the Cofunction Formula: sin(pi/2 - θ) = cos(θ)

which means: sin²(pi/2 - x) = cos²( x)

Second:

Using the Difference of Two Angle Identities: sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

sin(pi/2-x)=sin(pi/2)cos(x)-cos(pi/2)sin(x)

Since: sin(pi/2) = 1 and cos(pi/2) =0

sin(pi/2)cos(x)-cos(pi/2)sin(x) = 1*cos(x)-0*sin(x) = cos(x)

==>Which concurs with the Cofunction Formula in the first option.

Thus again means: sin²(pi/2 - x) = cos²( x)

Answer 3

Use cofunction identities, in this case take advantage of the fact that

cos((π/2) - x) = sin x,

sec((π/2) - x) = csc x,

therefore your problem can be rewritten as

sin x * (csc x - sin x)

= sin x * [(1/sin x) - sin x]

= sin x * [(1-sin^2 x)/sin x]

= 1 - sin^2 x

If you recall the Pythagorean identity for 1, then you will see this is

cos^2 x