Unit 3 || Trigonometry || Sin(a + b) Formula & Proof (Geometrical Explanation) || IB Math Guide

Proof of sin(A + B) Using Geometrical Construction

Result to Prove

\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]

Construction (Case 1: \(0^\circ < A+B \le 90^\circ\))

  • Let OX be the horizontal axis.
  • Construct angle \(A\), then extend to angle \(A+B\).
  • Take a point P on the terminal side of angle \(A+B\).
  • Draw \(PA \perp OX\).
  • Draw \(PB \perp\) terminal side of angle \(A\).
  • From B, draw \(BC \perp OX\).
  • From B, draw \(BD \perp PA\).
  • Join OB and OP.
P B A O

Step 1: Express sin(A + B)

\[ \sin(A+B) = \frac{AP}{OP} \]

Step 2: Split the Height

\[ AP = AD + DP \]

\[ \sin(A+B) = \frac{AD + DP}{OP} = \frac{AD}{OP} + \frac{DP}{OP} \]

Step 3: Express Each Term

\[ \frac{AD}{OP} = \frac{CB}{OB} \cdot \frac{OB}{OP} \]

\[ \frac{DP}{OP} = \frac{DP}{BP} \cdot \frac{BP}{OP} \]

Step 4: Convert into Trigonometric Ratios

  • \(\frac{CB}{OB} = \sin A\)
  • \(\frac{OB}{OP} = \cos B\)
  • \(\frac{DP}{BP} = \cos A\)
  • \(\frac{BP}{OP} = \sin B\)

Step 5: Final Substitution

\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]

Conclusion

Thus, using geometrical construction and decomposition of lengths, we have proven:

\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]

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