Exercise 13 Page 498

We want to write $cos 5 9^{\circ}$ in terms of sine. To do so, let's recall an important property of the sine and cosine of complementary angles.

The cosine of an acute angle is equal to the sine of its complement.

Keep in mind that complementary angles add up to

9 0^{\circ} .

To find the complement of

59,

we subtract our given angle from

90 .

Now, we can rewrite

cos 5 9^{\circ}

in terms of sine.

cos 5 9^{\circ} = sin (9 0^{\circ} - 5 9^{\circ}) = sin 3 1^{\circ}