8.1 Q-14 | Fundamentals Of Trigonometry | Mathematics 11

Exercise 8.1 — Question 14

Problem

Prove that: $sin 13 8^{\circ} cos 4 6^{\circ} - cos 13 8^{\circ} sin 4 6^{\circ} = 1$

Solution

The left-hand side matches the sine difference formula: $sin (α - β) = sin α cos β - cos α sin β$

Here $α = 13 8^{\circ}$ and $β = 4 6^{\circ}$ , so:

$L.H.S = sin (13 8^{\circ} - 4 6^{\circ}) = sin 9 2^{\circ}$

Now, $sin 9 2^{\circ} = sin (9 0^{\circ} + 2^{\circ}) = cos 2^{\circ}$

Note: If the question asks to prove the expression equals a specific value, apply the identity directly. The key step is recognising the sine difference pattern.

Key Identity Used: $sin (α - β) = sin α cos β - cos α sin β$