Answer:
Part 5.1.1:

Part 5.1.2:

Step-by-step explanation:
We are given that:

Part 5.1.1
Recall that:

Let θ = 2<em>A</em>. Hence:

Square the original equation:

Hence:

Subtract:

Take the square root of both sides:

Since 0° ≤ 2<em>A</em> ≤ 90°, cos(2<em>A</em>) must be positive. Hence:

Part 5.1.2
Recall that:

We can use the third form. Substitute:

Solve for cosine:

In conclusion:

(Note that since 0° ≤ 2<em>A</em> ≤ 90°, 0° ≤ <em>A</em> ≤ 45°. Hence, cos(<em>A</em>) must be positive.)
Answer:
Step-by-step explanation:
Three eights
-3x + 6y + 5 = -7
<u> -5 -5</u>
-3x + 6y = -12
-3x + 3x + 6y = -12 + 3x
<u>6y</u> = <u>-12 + 3x</u>
6 6
y = -2 + 1/2x
-3x + 6(-2 + 1/2x) = -12
-3x - 12 + 3x = -12
-3 + 3x - 12 = -12
0x - 12 = -12
<u> +12 +12</u>
<u>0x</u> = <u>0</u>
0 0
x = 0
-3(0) + 6y = -12
0 + 6y = -12
<u>+0 +0</u>
<u>6y</u> = <u>-12</u>
6 6
y = -2
(x, y) = (0, -2)