Why sin (A + B) is not sin A + sin B.

The rotated-triangle-in-a-rectangle picture: angle A stacked on angle B with hypotenuse 1, so the rectangle's sides read sin A cos B + cos A sin B and the height of the tip is sin (A + B). A side bar compares sin A + sin B.

Set A = 60° and B = 45° and read sin 105° from the height; then find the setting where sin A + sin B equals sin (A + B) (one angle 0°).

Writing sin (A + B) = sin A + sin B.

• cos (A + B) = cos A cos B − sin A sin B. It is proved on the unit circle: the chord joining the points at angles A and −B, and the chord joining the points at angles A + B and 0, subtend the same angle A + B at the centre, so they are equal, and equating their squared lengths gives the result.
• Putting −B for B gives cos (A − B) = cos A cos B + sin A sin B.
• With A = π/2 and B = θ, the difference formula gives cos (π/2 − θ) = 0 × cos θ + 1 × sin θ = sin θ; replacing θ by π/2 − θ then gives sin (π/2 − θ) = cos θ.
• sin (A + B) = sin A cos B + cos A sin B, and with −B for B, sin (A − B) = sin A cos B − cos A sin B.
• Signs to remember: in the cosine formulas the sign in the middle is the opposite of the sign inside the bracket; in the sine formulas it is the same.
• Worked values: cos 105° = cos (60° + 45°) = (1/2)(1/√2) − (√3/2)(1/√2) = (1 − √3)/(2√2), which is negative as a second-quadrant cosine must be, and sin 105° = sin (60° + 45°) = (√3/2)(1/√2) + (1/2)(1/√2) = (√3 + 1)/(2√2).
• These are identities, true for all real A and B. In general sin (A + B) ≠ sin A + sin B: at A = B = π/4 the left side is sin (π/2) = 1, but the right side is 2 × 1/√2 = √2.

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