8.1 Q-5 | Fundamentals Of Trigonometry | Mathematics 11

Exercise 8.1 — Question 5

Simplify using trigonometric identities:

$sin 13 8^{\circ} cos 4 6^{\circ} - cos 13 8^{\circ} sin 4 6^{\circ}$

The sine difference formula states:

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

Recognise the pattern $sin A cos B - cos A sin B$ with $α = 13 8^{\circ}$ and $β = 4 6^{\circ}$ :

$sin 13 8^{\circ} cos 4 6^{\circ} - cos 13 8^{\circ} sin 4 6^{\circ} = sin (13 8^{\circ} - 4 6^{\circ})$

$= sin 9 2^{\circ}$

Since $sin 9 2^{\circ} = sin (18 0^{\circ} - 8 8^{\circ}) = sin 8 8^{\circ}$ , or equivalently $sin 9 2^{\circ} \approx 0.9994$ .

$sin 9 2^{\circ}$

For any angles $α$ and $β$ :

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

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

These are deductions of the Fundamental Law of Trigonometry and are essential for simplifying compound-angle expressions.

Simplify using trigonometric identities:

$sin 13 8^{\circ} cos 4 6^{\circ} - cos 13 8^{\circ} sin 4 6^{\circ}$

The sine difference formula states:

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

Recognise the pattern $sin A cos B - cos A sin B$ with $α = 13 8^{\circ}$ and $β = 4 6^{\circ}$ :

$sin 13 8^{\circ} cos 4 6^{\circ} - cos 13 8^{\circ} sin 4 6^{\circ} = sin (13 8^{\circ} - 4 6^{\circ})$

$= sin 9 2^{\circ}$

Since $sin 9 2^{\circ} = sin (18 0^{\circ} - 8 8^{\circ}) = sin 8 8^{\circ}$ , or equivalently $sin 9 2^{\circ} \approx 0.9994$ .

$sin 9 2^{\circ}$

For any angles $α$ and $β$ :

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

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

These are deductions of the Fundamental Law of Trigonometry and are essential for simplifying compound-angle expressions.