8.1 Q-2 | Fundamentals Of Trigonometry | Mathematics 11

Exercise 8.1 — Question 2

Problem

Simplify each of the following using sum/difference identities:

(i) $sin 4 5^{\circ} cos 1 5^{\circ} + cos 4 5^{\circ} sin 1 5^{\circ}$

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

(iii) $cos 7 5^{\circ} cos 1 5^{\circ} + sin 7 5^{\circ} sin 1 5^{\circ}$

(iv) $cos 3 6^{\circ} cos 5 4^{\circ} - sin 3 6^{\circ} sin 5 4^{\circ}$

Key Identities Used

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

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

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

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

Solutions

(i) $sin 4 5^{\circ} cos 1 5^{\circ} + cos 4 5^{\circ} sin 1 5^{\circ}$

This matches the pattern $sin α cos β + cos α sin β = sin (α + β)$ .

$= sin (4 5^{\circ} + 1 5^{\circ}) = sin 6 0^{\circ} = \frac{3}{2}$

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

This matches the pattern $sin α cos β - cos α sin β = sin (α - β)$ .

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

Since $sin 9 2^{\circ} = sin (18 0^{\circ} - 8 8^{\circ}) = sin 8 8^{\circ}$ , or equivalently $sin 9 2^{\circ} \approx 1$ .

(iii) $cos 7 5^{\circ} cos 1 5^{\circ} + sin 7 5^{\circ} sin 1 5^{\circ}$

This matches the pattern $cos α cos β + sin α sin β = cos (α - β)$ .

$= cos (7 5^{\circ} - 1 5^{\circ}) = cos 6 0^{\circ} = \frac{1}{2}$

(iv) $cos 3 6^{\circ} cos 5 4^{\circ} - sin 3 6^{\circ} sin 5 4^{\circ}$

This matches the pattern $cos α cos β - sin α sin β = cos (α + β)$ .

$= cos (3 6^{\circ} + 5 4^{\circ}) = cos 9 0^{\circ} = 0$

Exercise 8.1 — Question 2

Problem

Simplify each of the following using sum/difference identities:

(i) $sin 4 5^{\circ} cos 1 5^{\circ} + cos 4 5^{\circ} sin 1 5^{\circ}$

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

(iii) $cos 7 5^{\circ} cos 1 5^{\circ} + sin 7 5^{\circ} sin 1 5^{\circ}$

(iv) $cos 3 6^{\circ} cos 5 4^{\circ} - sin 3 6^{\circ} sin 5 4^{\circ}$

Key Identities Used

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

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

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

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

Solutions

(i) $sin 4 5^{\circ} cos 1 5^{\circ} + cos 4 5^{\circ} sin 1 5^{\circ}$

This matches the pattern $sin α cos β + cos α sin β = sin (α + β)$ .

$= sin (4 5^{\circ} + 1 5^{\circ}) = sin 6 0^{\circ} = \frac{3}{2}$

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

This matches the pattern $sin α cos β - cos α sin β = sin (α - β)$ .

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

Since $sin 9 2^{\circ} = sin (18 0^{\circ} - 8 8^{\circ}) = sin 8 8^{\circ}$ , or equivalently $sin 9 2^{\circ} \approx 1$ .

(iii) $cos 7 5^{\circ} cos 1 5^{\circ} + sin 7 5^{\circ} sin 1 5^{\circ}$

This matches the pattern $cos α cos β + sin α sin β = cos (α - β)$ .

$= cos (7 5^{\circ} - 1 5^{\circ}) = cos 6 0^{\circ} = \frac{1}{2}$

(iv) $cos 3 6^{\circ} cos 5 4^{\circ} - sin 3 6^{\circ} sin 5 4^{\circ}$

This matches the pattern $cos α cos β - sin α sin β = cos (α + β)$ .

$= cos (3 6^{\circ} + 5 4^{\circ}) = cos 9 0^{\circ} = 0$