Mixing
find theta from sin theta
$begingroup$
having problems on these trig questions, not really sure how to go about them. If you could tell me the method to go about it would be great, thanks!
1
a) Find $theta$ such that $sin(theta) = sin(99pi/5) quad text{and} quad -frac {1}2 pi leq theta leq frac 12 pi$
b) Find $theta$ such that $cos(theta) = cos(-94pi/7) quad text{and} quad 0pi leq theta leq pi$
2
Suppose $x$ is in the third quadrant, and $sin x = -1/3$. Find, without using any of the trig capabilities of your calculator, each of the following:
a) $cos x$
b) $sin 2x$
c) $cos 2x$
d) $sinleft(dfrac{x}{2}right)$
trigonometry
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
$endgroup$
add a comment |
$begingroup$
having problems on these trig questions, not really sure how to go about them. If you could tell me the method to go about it would be great, thanks!
1
a) Find $theta$ such that $sin(theta) = sin(99pi/5) quad text{and} quad -frac {1}2 pi leq theta leq frac 12 pi$
b) Find $theta$ such that $cos(theta) = cos(-94pi/7) quad text{and} quad 0pi leq theta leq pi$
2
Suppose $x$ is in the third quadrant, and $sin x = -1/3$. Find, without using any of the trig capabilities of your calculator, each of the following:
a) $cos x$
b) $sin 2x$
c) $cos 2x$
d) $sinleft(dfrac{x}{2}right)$
trigonometry
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
$endgroup$
2
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09
add a comment |
$begingroup$
having problems on these trig questions, not really sure how to go about them. If you could tell me the method to go about it would be great, thanks!
1
a) Find $theta$ such that $sin(theta) = sin(99pi/5) quad text{and} quad -frac {1}2 pi leq theta leq frac 12 pi$
b) Find $theta$ such that $cos(theta) = cos(-94pi/7) quad text{and} quad 0pi leq theta leq pi$
2
Suppose $x$ is in the third quadrant, and $sin x = -1/3$. Find, without using any of the trig capabilities of your calculator, each of the following:
a) $cos x$
b) $sin 2x$
c) $cos 2x$
d) $sinleft(dfrac{x}{2}right)$
trigonometry
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
$endgroup$
having problems on these trig questions, not really sure how to go about them. If you could tell me the method to go about it would be great, thanks!
1
a) Find $theta$ such that $sin(theta) = sin(99pi/5) quad text{and} quad -frac {1}2 pi leq theta leq frac 12 pi$
b) Find $theta$ such that $cos(theta) = cos(-94pi/7) quad text{and} quad 0pi leq theta leq pi$
2
Suppose $x$ is in the third quadrant, and $sin x = -1/3$. Find, without using any of the trig capabilities of your calculator, each of the following:
a) $cos x$
b) $sin 2x$
c) $cos 2x$
d) $sinleft(dfrac{x}{2}right)$
trigonometry
trigonometry
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
edited Mar 22 '16 at 11:17
N. F. Taussig
44.5k103357
44.5k103357
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
asked Mar 22 '16 at 7:11
TrigamatuerTrigamatuer
4716
4716
2
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09
add a comment |
2
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09
2
2
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For question #1 you need to use the formula:
$$sin theta = sin aimplies theta =begin {cases} 2 k pi + a,& k in mathbb Z \\
text{or}\\
2kpi +pi - a,&k in mathbb Z end{cases}$$
and in the second case
$$cos theta = cos aimplies theta = 2kpi pm a, quad k in mathbb Z$$
In our example, in the first case $a = 99pi/5.$ Take advantage of the inequality to find all the appropriate $kin mathbb Z$ in order to define $theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$sin^2 x=frac{1-cos 2x}{2}$$
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For question #1 you need to use the formula:
$$sin theta = sin aimplies theta =begin {cases} 2 k pi + a,& k in mathbb Z \\
text{or}\\
2kpi +pi - a,&k in mathbb Z end{cases}$$
and in the second case
$$cos theta = cos aimplies theta = 2kpi pm a, quad k in mathbb Z$$
In our example, in the first case $a = 99pi/5.$ Take advantage of the inequality to find all the appropriate $kin mathbb Z$ in order to define $theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$sin^2 x=frac{1-cos 2x}{2}$$
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
$endgroup$
add a comment |
$begingroup$
For question #1 you need to use the formula:
$$sin theta = sin aimplies theta =begin {cases} 2 k pi + a,& k in mathbb Z \\
text{or}\\
2kpi +pi - a,&k in mathbb Z end{cases}$$
and in the second case
$$cos theta = cos aimplies theta = 2kpi pm a, quad k in mathbb Z$$
In our example, in the first case $a = 99pi/5.$ Take advantage of the inequality to find all the appropriate $kin mathbb Z$ in order to define $theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$sin^2 x=frac{1-cos 2x}{2}$$
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
$endgroup$
add a comment |
$begingroup$
For question #1 you need to use the formula:
$$sin theta = sin aimplies theta =begin {cases} 2 k pi + a,& k in mathbb Z \\
text{or}\\
2kpi +pi - a,&k in mathbb Z end{cases}$$
and in the second case
$$cos theta = cos aimplies theta = 2kpi pm a, quad k in mathbb Z$$
In our example, in the first case $a = 99pi/5.$ Take advantage of the inequality to find all the appropriate $kin mathbb Z$ in order to define $theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$sin^2 x=frac{1-cos 2x}{2}$$
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
$endgroup$
For question #1 you need to use the formula:
$$sin theta = sin aimplies theta =begin {cases} 2 k pi + a,& k in mathbb Z \\
text{or}\\
2kpi +pi - a,&k in mathbb Z end{cases}$$
and in the second case
$$cos theta = cos aimplies theta = 2kpi pm a, quad k in mathbb Z$$
In our example, in the first case $a = 99pi/5.$ Take advantage of the inequality to find all the appropriate $kin mathbb Z$ in order to define $theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$sin^2 x=frac{1-cos 2x}{2}$$
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
edited Mar 22 '16 at 8:42
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
answered Mar 22 '16 at 7:41
thanasissdrthanasissdr
5,55111325
5,55111325
add a comment |
add a comment |
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2
$begingroup$
Start simplifying the rhs : $frac{99pi}{5}=20pi-frac pi 5$ and $frac{94pi}{7}=13pi+frac{3 pi}7$
$endgroup$
– Claude Leibovici
Mar 22 '16 at 8:09