Mixing
Find $sin(theta)$ when $sin(theta) > 0$, $tan(theta)=7/24$.
$begingroup$
Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.
I have no idea what to do.
trigonometry
edited Jan 18 at 12:21
Yanko
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
$endgroup$
add a comment |
$begingroup$
Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.
I have no idea what to do.
trigonometry
edited Jan 18 at 12:21
Yanko
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
$endgroup$
3
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24
add a comment |
$begingroup$
Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.
I have no idea what to do.
trigonometry
edited Jan 18 at 12:21
Yanko
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
$endgroup$
Find $sin(theta)$ when $sin(theta) > 0$,
$tan(theta)=7/24$.
I have no idea what to do.
trigonometry
trigonometry
edited Jan 18 at 12:21
Yanko
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
edited Jan 18 at 12:21
Yanko
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
edited Jan 18 at 12:21
Yanko
7,1751729
edited Jan 18 at 12:21
Yanko
7,1751729
edited Jan 18 at 12:21
Yanko
7,1751729
7,1751729
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
asked Jan 18 at 12:18
Renee HaasRenee Haas
11
11
3
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24
add a comment |
3
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24
3
3
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.
The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
$endgroup$
add a comment |
$begingroup$
Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint
use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
use $$sin theta ={tanthetaover sqrt{1+tan^2theta}}$$whenever $sin theta , tan theta ge 0$.
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 18 at 12:23
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.
The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
$endgroup$
add a comment |
$begingroup$
Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.
The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
$endgroup$
add a comment |
$begingroup$
Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.
The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
$endgroup$
Draw a right triangle that shows $tan theta = 7/24.$ There are infinitely many, but choosing the one with legs $7$ and $24$ is a swell choice. Now use Pythagorean Theorem to find the length of the hypotenuse. Finding the values of any of the other trig functions should be easy now.
The only snag is that you have to think about which quadrant you're in. But you have $sin theta$ and $tan theta$ both positive, so you're in the first quadrant.
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
answered Jan 18 at 12:28
B. GoddardB. Goddard
19.1k21441
19.1k21441
add a comment |
add a comment |
$begingroup$
Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
$endgroup$
Use $cos^2theta = 1 - sin^2theta$ and $tantheta = frac{sintheta}{costheta}=c$. Then $sintheta = csqrt{1-sin^2theta}$, then $sin^2theta = c^2-c^2sin^2theta$, and finally $sintheta = frac{c}{sqrt{1+c^2}}$. Note that when taking the roots I have taken the positive such since $sintheta>0$ and also $tantheta >0$, which implies $costheta > 0$.
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
edited Jan 18 at 12:34
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
answered Jan 18 at 12:27
lightxbulblightxbulb
945311
945311
add a comment |
add a comment |
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3
$begingroup$
You're given the ratio between sine and cosine. Also the sum of their squares is one...
$endgroup$
– Matteo
Jan 18 at 12:21
$begingroup$
Hint: $sec^2(x)=1+tan^2(x)$ and $sin(x)=frac{tan(x)}{sec(x)}$
$endgroup$
– robjohn♦
Jan 18 at 12:24