{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Pre-Calc Exam Notes 80

# Pre-Calc Exam Notes 80 - same sign Thus the minus sign in...

This preview shows page 1. Sign up to view the full content.

80 Chapter 3 Identities §3.3 By taking square roots, we can write the above formulas in an alternate form: sin 1 2 θ = ± radicalBigg 1 cos θ 2 (3.31) cos 1 2 θ = ± radicalBigg 1 + cos θ 2 (3.32) tan 1 2 θ = ± radicalBigg 1 cos θ 1 + cos θ (3.33) In the above form, the sign in front of the square root is determined by the quadrant in which the angle 1 2 θ is located. For example, if θ = 300 then 1 2 θ = 150 is in QII. So in this case cos 1 2 θ < 0 and hence we would have cos 1 2 θ =− radicalBig 1 + cos θ 2 . In formula (3.33), multiplying the numerator and denominator inside the square root by (1 cos θ ) gives tan 1 2 θ = ± radicalBigg 1 cos θ 1 + cos θ · 1 cos θ 1 cos θ = ± radicalBigg (1 cos θ ) 2 1 cos 2 θ = ± radicalBigg (1 cos θ ) 2 sin 2 θ = ± 1 cos θ sin θ . But 1 cos θ 0, and it turns out (see Exercise 10) that tan
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: same sign. Thus, the minus sign in front of the last expression is not possible (since that would switch the signs of tan 1 2 θ and sin θ ), so we have: tan 1 2 θ = 1 − cos θ sin θ (3.34) Multiplying the numerator and denominator in formula (3.34) by 1 + cos θ gives tan 1 2 θ = 1 − cos θ sin θ · 1 + cos θ 1 + cos θ = 1 − cos 2 θ sin θ (1 + cos θ ) = sin 2 θ sin θ (1 + cos θ ) , so we also get: tan 1 2 θ = sin θ 1 + cos θ (3.35) Taking reciprocals in formulas (3.34) and (3.35) gives: cot 1 2 θ = sin θ 1 − cos θ = 1 + cos θ sin θ (3.36)...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online