How Are the Trigonometric Identities for Cosine and Sine Related?

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nesta
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Hi friends,

Please make me understand this simplest function,

y = cos θ

2. in the next step it says: cos θ = sin (π/2 - θ)
3. and similarly -cos (π/2 - θ) = -sin θ.

Can anyone please explain how the steps 2 & 3 are deduced.

Thanks,
Nesta
 
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nesta said:
Hi friends,

Please make me understand this simplest function,

y = cos θ

2. in the next step it says: cos θ = sin (π/2 - θ)
3. and similarly -cos (π/2 - θ) = -sin θ.

Can anyone please explain how the steps 2 & 3 are deduced.

Thanks,
Nesta
The most basic definition of "cosine" is that it is "near side divided by hypotenuse" in a right triangle and of "sine" that it is "opposite side divided by hypotenuse".
Since a right triangle has one angle of size 90 degrees or [itex]\pi/2[/itex] radians, and the angles in any triangle sum to [itex]\pi[/itex] radians, the two acute angles must sum to [itex]\pi- \pi/2= \pi/2[/itex]. That is, if one of the acute angles is [itex]\theta[/itex], then the other is [itex]\pi/2- \theta[/itex]. And, of course, switching angles swaps "near" and "opposite" sides.

For a more general definition, where [itex]\theta[/itex] is not restricted to be between 0 and [itex]\pi/2[/itex] radians, you would have to go with something like the unit circle definition rochfor1 suggests.