Answer:
The work is in the explanation.
Step-by-step explanation:
The sine addition identity is:
.
The sine difference identity is:
.
The cosine addition identity is:
.
The cosine difference identity is:
.
We need to find a way to put some or all of these together to get:
.
So I do notice on the right hand side the 
and the 
.
Let's start there then.
There is a plus sign in between them so let's add those together:
![=[\sin(a+b)]+[\sin(a-b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%2Bb%29%5D%2B%5B%5Csin%28a-b%29%5D)
![=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Ccos%28a%29%5Csin%28b%29%5D%2B%5B%5Csin%28a%29%5Ccos%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)
There are two pairs of like terms. I will gather them together so you can see it more clearly:
![=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Csin%28a%29%5Ccos%28b%29%5D%2B%5B%5Ccos%28a%29%5Csin%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)
So this implies:
Divide both sides by 2:
By the symmetric property we can write:
I think this is what you are looking for:
=−12
a
b
=
−
12
+=4
a
+
b
=
4
(+)2=42
(
a
+
b
)
2
=
4
2
2+2+2=16
a
2
+
b
2
+
2
a
b
=
16
∴2+2=16+2×12=40
∴
a
2
+
b
2
=
16
+
2
×
12
=
40
Now, (−)2=2+2−2=40+2×12=64
(
a
−
b
)
2
=
a
2
+
b
2
−
2
a
b
=
40
+
2
×
12
=
64
∴(−)=64‾‾‾√=±8
∴
(
a
−
b
)
=
64
=
±
8
So, 2−2=(+)(−)
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
2−2=(4)(±8)=±32
a
2
−
b
2
=
(
4
)
(
±
8
)
=
±
32
Hope that helps and sorry if it is confusing!
1/12 does not belong with the other three fractions.
2/12, 3/12, and 4/12 can all be simplified to 1/6, 1/4/, and 1/3. However, 1/12 can't be simplified.
We know:
We have
Substitute:
Only 4 and 10 meet the requirements of the question.
--------------------------------------------------------------------
40 and 1 NOT, because 40 > 12
20 and 2 NOT, because 20 > 12
5 and 8 NOT because GCF(5, 8) = 1
Answer:it ends at 11:00
Step-by-step explanation:8:30+2:30=11:00
