Answer:
Part a
See proof below
Part b
Whenever tan(α) = 1
So at α = 45° or α = 225°
Explanation:
Part a
Since α,β ae complementary angles,
α = 90 - β and β = 90 - α
Use the expansion:
cos(A-B)=cos (A) · cos (B) + sin(A) · sin(B)
cos(β) = cos(90 - α) = cos(90) · cos(α) + sin(90)· sin(α)
cos(90) = 0 and sin(90) = 1 which results in
cos(β) = 0· cos(α) + 1 · sin(α)
cos(β) = sin(α)
or
sin(α) = cos(β)
In a similar manner
cos(α) = cos(90-β)
= cos(90) · cos(β) + sin(90 ) · sin(β)
= 0 · cos(β) + 1 · sin(β)
= sin(β)
cos(α) = sin(β) or in other words,
sin(β) = cos(α)
b. sin(α) = cos(α) when α = 45° or α = 135°
If sin(α) = cos(α) then
tan(α) = sin(α)/cos(α) = 1
α = tan⁻¹(α) = 45° or at α = 225°
So wherever tan(α) is 1 we will have sin(α) = cos(α)
This occurs at two locations - in quadrants 1 and 3
Quadrant 1
cos(45) = 1/√2
and
sin(45) = 1/√2
Quadrant 3
sin(225) = - √2/2
cos(225) = - √2/2