Property of sine function and cosine function?

Question

My teacher wrote that if theta one is any acute value such that sin of theta = k, then another value of theta which satisfies this is theta 2 = pi-theta one. 

 

And for cosine it is theta2=2pi-theta one.

 

What does this mean?

Answer 1

Edward,      Θ

 

First notice that the Theta angles are specified in radians, not degrees.  One π (Pi) radians is really 180^o and 2π radians is the full circle or 360^o.

 

Imagine the acute angle Θ₁ in quadrant I, drop a perpendicular to the x-axis through x=1 and you have a right triangle whose height is sin(Θ1) = k and whose base is 1.  Now rotate this triangle about the y-axis into quadrant II.  None of the angles or the lengths inside that triangle have changed.  Θ₂ is now the angle measured from the +x-axis to the hypotenuse of that rotated triangle.  So   Θ₂ = ( π - Θ₁) radians.  And  sin(Θ₂) = sin(Θ₁) = k  by construction.

 

To prove this analytically, start with the known relationship:

sin (x - y) = sin (x) cos (y) - cos(x) sin(y),  substitute in x = π   and  y = +Θ₁  

sin (π - Θ₁) = sin (π) cos (Θ₁) - cos(π) sin(Θ₁), and note that sin (π) = 0  and cos(π) = -1 in quadrant II

Thus, sin (π - Θ₁) =  - (-1) sin(Θ₁) = sin(Θ₁) = k

 

For the second part, rotate the quadrant I triangle about the x-axis into Quadrant IV.

Now Θ₂ = ( 2π - Θ₁)  because its the angle that starts at the +x-axis, goes through quadrants I, II, III into IV and ends Θ₁ radians shy of the x-axis  or 2π.  And cos (Θ₁) = 1 by construction in quadrants I & IV.

 

To prove this analytically, start with another known relationship:
cos (x - y) = cos (x) cos (y) - sin (x) sin (y), substitute in x = 2π and y = +Θ₁
cos (2π - Θ₁) = cos (2π) cos (Θ₁) - sin (2π) sin(Θ₁), and note that sin (2π) = 0 and cos(2π) = +1 in quadrant IV
Thus, cos (2π - Θ1) = (+1) cos (Θ₁) = cos (Θ1) = 1

Answer 2

Consider any value of θ in the domain 0 ≤ θ₁ ≤ ^π⁄₂ (where θ₁ is acute or right).  If we take the distance from θ₁ to ^π⁄₂, add that difference to ^π⁄₂, and call that θ₂, then sin(θ₁) = sin(θ₂).  

 

This is just a way of saying that the sine function is symmetric about θ = ^π⁄₂ in the domain 0 ≤ θ₁ ≤ π. To test this, draw a vertical line through the graph of the sine function at θ = ^π⁄₂ and fold the page on that line; the function is reflected there.  Alternately, consider that ^π⁄₆ = ^π⁄₂ - ^π⁄₃, ^5π⁄₆ = ^π⁄₂ + ^π⁄₃, and sin(^π⁄₆) = sin(^5π⁄₆) = ¹/₂.

 

Since θ₂ = ^π⁄₂ + (^π⁄₂ - θ₁), we have θ₂ = ^π⁄₂ + (^π⁄₂ - θ₁) = π - θ1.

 

A similar principle is at work in the case of the cosine.