Download Math 1113 Solutions: Trigonometric Identities and Graphs and more Exams Pre-Calculus in PDF only on Docsity!
Roll No Name: Test # 2, Solutions Math 1113, October 8th, 2006
- Calculate cot(arccos(− 5 /13)).
Solution: Since cot(α) = cos sin αα and sin α = ±
1 − cos^2 (α) we obtain
cot(arccos(− 5 /13)) =
cos[arccos(− 5 /13)] sin[arccos(− 5 /13)]
1 − (− 5 /13)^2
where the sign needs to be determined by the argument that the angle arccos(− 5 /13) is in the second quadrant and so cotanget is
negative there. Hence, cot(arccos(− 5 /13)) = −
- Verify the identity:
tan a + tan b cot a + cot b
= tan a tan b.
Solution: Since cot α = (^) tan^1 α we have
LHS =
tan a + tan b 1 tan a +^
1 tan b
tan a + tan b tan a+tan b tan a tan b
tan a + tan b 1
tan a tan b tan a + tan b
= tan a tan b = RHS.
- Graph the functions f (x) = cos(x), g(x) = cos(π − x). What can you say about f and g?
Solution: The graph of f on the interval [− 4 π, 4 π] is included below:
1
0
-5^05 -0.
-10 x 10
and the graph of g on the same interval is 1
0
-5 0 5 -0.
-10 x 10
which suggests that if we flip one over the x-axis we get the other.
This means g(x) = −f (x) or cos(π − x) = − cos x for all x ∈ R.
- Calculate the exact value of cos(2 arcsin 3/4).
Solution: We have the double angle formula: cos(2α) = 1−2 sin^2 α. So
cos(2 arcsin 3/4) = 1−2 sin^2 (arcsin(3/4)) = 1−2(3/4)^2 = 1−2(
and so cos(2 arcsin 3/4) = −
- Verify the identity sin(α − β) sin(α + β) = sin^2 (α) − sin^2 (β).
Solution: We just the formulae sin(α−β) = sin α cos β −cos α sin β and sin(α + β) = sin α cos β + cos α sin β which gives
LHS = sin(α − β) sin(α + β) = sin^2 α cos^2 β − cos^2 α sin^2 β =
sin^2 α(1 − sin^2 β) − (1 − sin^2 α) sin^2 β = sin^2 (α) − sin^2 (β) = RHS.
- Check that
sin 3t sin t
cos 3t cos t
= 4 cos 2t.
Solution: Adding to the same common denominator we obtain:
LHS =
sin 3t sin t
cos 3t cos t
sin 3t cos t + cos 3t sin t sin t cos t
sin 4t sin t cos t
2 sin 2t cos 2t sin t cos t
4 sin t cos t cos 2t sin t cos t
= 4 cos 2t = RHS.
