Is not because the 13-3 would be positive and the other 3-13 would be negative
Answer:
![\sec(x)=\frac{1}{\sqrt{1-m^2}}](https://tex.z-dn.net/?f=%5Csec%28x%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-m%5E2%7D%7D)
Step-by-step explanation:
We know that:
![\sin(-x)=-m](https://tex.z-dn.net/?f=%5Csin%28-x%29%3D-m)
First, since sine is an odd function, we can move the negative outside:
![=-\sin(x)=-m](https://tex.z-dn.net/?f=%3D-%5Csin%28x%29%3D-m)
Divide both sides by -1:
![\sin(x)=m](https://tex.z-dn.net/?f=%5Csin%28x%29%3Dm)
We will now use the Pythagorean Identity:
![\cos^2(x)+\sin^2(x)=1](https://tex.z-dn.net/?f=%5Ccos%5E2%28x%29%2B%5Csin%5E2%28x%29%3D1)
Substitute m for sine:
![\cos^2(x)+m^2=1](https://tex.z-dn.net/?f=%5Ccos%5E2%28x%29%2Bm%5E2%3D1)
Solve for cosine:
![\cos^2(x)=1-m^2](https://tex.z-dn.net/?f=%5Ccos%5E2%28x%29%3D1-m%5E2)
Take the square root of both sides:
![\cos(x)=\pm\sqrt{1-m^2}](https://tex.z-dn.net/?f=%5Ccos%28x%29%3D%5Cpm%5Csqrt%7B1-m%5E2%7D)
Since x is an acute angle, cosine will always be positive. Thus:
![\cos(x)=\sqrt{1-m^2}](https://tex.z-dn.net/?f=%5Ccos%28x%29%3D%5Csqrt%7B1-m%5E2%7D)
Take the reciprocal of both sides. Hence:
![\frac{1}{\cos(x)}=\sec(x)=\frac{1}{\sqrt{1-m^2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Ccos%28x%29%7D%3D%5Csec%28x%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-m%5E2%7D%7D)
Answer:
they are opposites and parallel :)
Step-by-step explanation:
quizlet... XD
Answer:
3
Step-by-step explanation:
This is the graph of y = tan(x) shifted 3 units up.
Answer:
θ
∈
{
90
∘
,
210
∘
,
270
∘
,
330
∘
}
step by step explanation:
Factoring the left side of
2
sin
(
θ
)
cos
(
θ
)
+
cos
(
θ
)
=
0
gives
cos
(
θ
)
⋅
[
2
sin
(
θ
)
+
1
]
=
0
Which implies (for θ
∈
[
0
,
360
∘])
either cos
(
θ
)
=
0
→ θ
=
90
∘
or
270
∘
OR
2
sin
(
θ
)
+
1
=
0
→ sin (
θ
)=−
1
2
→
θ
=
210
∘
or
330
∘
→ θ = 210∘ or 330∘