8.1 Q-3 | Fundamentals Of Trigonometry | Mathematics 11

Exercise 8.1 — Question 3

Problem Statement

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 intended result is $sin (9 2^{\circ})$ , verify the exact problem statement from your textbook, as some versions of this question yield $sin 9 2^{\circ}$ rather than exactly $1$ .

Key Identity Used

$sin (α - β) = sin α cos β - cos α sin β$

Exercise 8.1 — Question 3

Problem Statement

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 intended result is $sin (9 2^{\circ})$ , verify the exact problem statement from your textbook, as some versions of this question yield $sin 9 2^{\circ}$ rather than exactly $1$ .

Key Identity Used

$sin (α - β) = sin α cos β - cos α sin β$