Mixed problems

Most problems in this chapter come down to three habits: look at the angles first, pick the identity that matches what you see, and fix every sign from the quadrant.

Look at the angles. Pairs adding to 90° or 180° often cancel or combine. Angles that double or halve each other call for the double-angle formulas. A product calls for product → sum; a sum calls for sum → product.

A product of sines. Show sin 10° sin 50° sin 70° = 1/8.

• 2 sin 50° sin 70° = cos 20° – cos 120° = cos 20° + 1/2.
• So the product is (1/2) sin 10° (cos 20° + 1/2) = (1/2) sin 10° cos 20° + (1/4) sin 10°.
• 2 sin 10° cos 20° = sin 30° – sin 10°, so the first part is (1/4)(1/2 – sin 10°).
• Adding: (1/4)(1/2) – (1/4) sin 10° + (1/4) sin 10° = 1/8.

Half angles. From 1 – cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2) you can find the ratios of x/2. Work out which quadrant x/2 is in first; it is usually different from x's quadrant.

Example: sin x = 4/5 with π/2 < x < π.

• cos x = –3/5 (quadrant II). Halving the range, π/4 < x/2 < π/2, so x/2 is in quadrant I and everything is positive.
• sin²(x/2) = (1 + 3/5)/2 = 4/5, so sin(x/2) = 2/√5. cos²(x/2) = (1 – 3/5)/2 = 1/5, so cos(x/2) = 1/√5.
• tan(x/2) = 2.

The same idea gives exact values for new angles: cos²(π/8) = (1 + cos π/4)/2 = (2 + √2)/4, so cos π/8 = √(2 + √2)/2 (positive, since π/8 is in quadrant I).

Square roots need care. √(2 + 2cos 2x) = √(4cos²x) = 2 × (the positive value of cos x). For 0 < x < π/2, cos x is positive, so it equals 2 cos x; for an angle where cos x is negative it would be –2 cos x.

Before you finish, check one easy value. For instance, put x = 0 or x = 30° into both sides of an identity you have proved: if they disagree, there is a slip somewhere.