Relation between Trigonometric Functions of Same Argument

We already know that `cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)`.

If we take `alpha=t,beta=t` then `cos(t-t)=cos(t)cos(t)+sin(t)sin(t)` or `cos(0)=cos^2(t)+sin^2(t)`.

This gives us very important identity that connects sine and cosine (it is also called main trigonometric identity): `cos^2(t)+sin^2(t)=1`.

Dividing main trigonometric identity by `cos^2(t)` gives that `1+tan^2(t)=1/(cos^2(t))=sec^2(t)`. Dividing main trigonometric identity by `sin^2(t)` gives that `1+cot^2(t)=1/(sin^2(t))=csc^2(t)`.

So, we have three formulas:

  1. `color(red)(cos^2(t)+sin^2(t)=1)`,
  2. `color(blue)(1+tan^2(t)=sec^2(t))`,
  3. `color(green)(1+cot^2(t)=csc^2(t))`.

Formula (2) holds when `t!=pi/2+pin, n in ZZ` (when `cos(t)!=0`).Formula (3) holds when `t!=pin, n in ZZ` (when `sin(t)!=0`).

These three formulas connect different trigonometric functions of same argument.

There are also two additional identities that connect different trigonometric functions of same argument: `tan(t)=(sin(t))/(cos(t)),cot(t)=(cos(t))/(sin(t))` . But in fact these identities are definitions of tangent and cotangent.

Multiplying above two identities, we will obtain relation `tan(t)xxcot(t)=1` that is true when `t!=(pi k)/2, k in ZZ` (when both `cos(t)!=0` and `sin(t)!=0`).

Example. It is known that `sin(t)=-3/5` and `pi<t<(3pi)/2`. Find `cos(t),tan(t),cot(t)`.

From formula (1) we have that `cos^2(t)=1-sin^2(t)=1-(-3/5)^2=1-9/25=16/25`.

Since `cos^2(t)=16/25`, then either `cos(t)=sqrt(16/25)=4/5` or `cos(t)=-sqrt(16/25)=-4/5`.

Since `pi<t<(3pi)/2`, then `t` belongs to third quadrant. In this quadrant cosine is negative, so `cos(t)=-4/5`.

Now, `tan(t)=(sin(t))/(cos(t))=(-3/5)/(-4/5)=3/4` and `cot(t)=1/(tan(t))=4/3`.

So, `cos(t)=-4/5,tan(t)=3/4,cot(t)=4/3`.