Build addition formulas from rotated coordinates

Try

Suppose an arrow has coordinates (u,v). Think of it as a horizontal displacement u followed by a vertical displacement v. Rotate the entire arrangement counterclockwise through angle a. The two component directions rotate with it, so adding their rotated displacements must give the rotated arrow.

This construction derives trigonometric formulas rather than assuming them. You need signed coordinates, the unit circle and basic geometry of a rigid rotation; matrix notation is optional. Negative components point opposite their named positive direction.

Predict

The horizontal unit direction (1,0) rotates to (cos a,sin a). Predict where the vertical unit direction (0,1) goes. It remains perpendicular and is a quarter-turn counterclockwise from the first direction, giving (−sin a,cos a). Check a=0 and a=π/2 before using the general rule.

Experiment

The dashed rays show perpendicular unit directions after rotation. The thick path first travels along a scaled U and then along a scaled V; its endpoint is their sum. For the displayed example cos a=4/5 and sin a=3/5, so U=(4/5,3/5) and V=(−3/5,4/5). The component lengths are 5/13 and 12/13, giving endpoint (−16/65,63/65). These are exact mathematical coordinates; the drawing uses the same scale horizontally and vertically.

Draw a unit square’s two directed sides and rotate both by an acute angle a. Label each signed horizontal and vertical projection. Now scale and add the directions as shown.

Original displacement	Rotated horizontal part	Rotated vertical part
u times (1,0)	u cos a	u sin a
v times (0,1)	−v sin a	v cos a
Combined (u,v)	u cos a−v sin a	u sin a+v cos a

Take (cos a,sin a)=(4/5,3/5) and rotate (u,v)=(5/13,12/13). The result is (−16/65,63/65): horizontal part 20/65−36/65 and vertical part 15/65+48/65. Verify its squared length is one. Its negative horizontal coordinate is possible even though both input angles are acute: their sum can exceed π/2.

Understand

Why can rotated components be added? A rigid rotation about the origin preserves the parallelogram construction for vector addition and preserves directed lengths along each ray. Rotating a scaled direction therefore scales its rotated direction by the same signed amount. This is a geometric property of rotation, not an already assumed trigonometric addition identity.

Set u=cos b and v=sin b in the table. The original arrow ends at angle b; rotation by a ends at angle a+b. Equating its coordinates gives

cos(a+b)=cos a cos b−sin a sin b,

sin(a+b)=sin a cos b+cos a sin b.

Both identities hold for all real angles. They are not limited to acute triangles because signed component decomposition works in every quadrant. To obtain subtraction, replace b with −b. Reflection gives cos(−b)=cos b and sin(−b)=−sin b. Consequently cosine subtraction has a plus between its products, while sine subtraction has a minus:

cos(a−b)=cos a cos b+sin a sin b,

sin(a−b)=sin a cos b−cos a sin b.

Tangent requires more care. Take the sine-addition formula divided by cosine-addition. Dividing both numerator and denominator by cos a cos b produces

tan(a+b)=(tan a+tan b)/(1−tan a tan b).

This rational expression requires cos a≠0, cos b≠0 and 1−tan a tan b≠0. The last restriction ensures the combined angle also has nonzero cosine. Similarly, subtraction gives (tan a−tan b)/(1+tan a tan b), with both individual tangents defined and that denominator nonzero. Deriving a formula by division does not extend its domain to inputs excluded by the division.

For a=b=π/2, the angle a+b equals π and its tangent is zero, but neither individual tangent exists. The rational expression is unusable there. Conversely, when a=b=π/4, the individual tangents exist but their addition denominator is zero, correctly warning that tan(π/2) is undefined.

Apply

Set a=b=x to derive sin(2x)=2sin x cos x. Use it to solve sin(2x)=0 on [0,2π]. The factors give sine zeros and cosine zeros, so the solutions are 0,π/2,π,3π/2,2π. Dividing by either factor before checking its zeros would lose answers. Verify the endpoints as actual points of the closed interval.

Continue

Rebuild one formula from the rotated directions without looking at the finished expressions. Then use existing sum and double-angle practice for more applications. The next lesson explains why reversing a trigonometric calculation needs a restricted domain rather than a single inverse key that returns every angle.