103 Trigonometry Problems

4. Solutions to Introductory Problems 85
(d) Applying the double-angle formulas again gives
8 sin 20 ◦ sin 10 ◦ sin 50 ◦ sin 70 ◦ = 8 sin 20 ◦ cos 20 ◦ cos 40 ◦ cos 80 ◦
= 4 sin 40 ◦ cos 40 ◦ cos 80 ◦
= 2 sin 80 ◦ cos 80 ◦
= sin 160 ◦ = sin 20 ◦ .
Consequently,
sin 10 ◦ sin 50 ◦ sin 70 ◦ = 1 8 .
4. [AMC12P 2002] Simplify the expression
√
√
sin 4 x + 4 cos 2 x − cos 4 x + 4 sin 2 x.
Solution: The given expression is equal to
√
sin 4 x + 4(1 − sin 2 x) − √ cos 4 x + 4(1 − cos 2 x)
√
= (2 − sin 2 x) 2 − √ (2 − cos 2 x) 2 = (2 − sin 2 x) − (2 − cos 2 x)
= cos 2 x − sin 2 x = cos 2x.
5. Prove that
1 − cot 23 ◦ =
2
1 − cot 22 ◦ .
First Solution: We will show that
(1 − cot 23 ◦ )(1 − cot 22 ◦ ) = 2.
Indeed, by the addition and subtraction formulas, we obtain
)( )
(1 − cot 23 ◦ )(1 − cot 22 ◦ cos 23◦ cos 22◦
) =
(1 −
sin 23 ◦ 1 −
sin 22 ◦
= sin 23◦ − cos 23 ◦
sin 23 ◦ · sin 22◦ − cos 22 ◦
√
sin 22 ◦
2 sin(23 ◦ − 45 ◦ ) √ 2 sin(22 ◦ − 45 ◦ )
=
sin 23 ◦ · sin 22 ◦
= 2 sin(−22◦ ) sin(−23 ◦ )
sin 23 ◦ sin 22 ◦
= 2 sin 22◦ sin 23 ◦
sin 23 ◦ sin 22 ◦ = 2.