What is the equation of the line that passes through the points (0, 6) and (-3, 0) and is parallel to y = 2x + 5? *

Mathematics

1 answer:

adoni [48]3 years ago

4 0

Answer:

Step-by-step explanation:

Using the slope from the other equation, and the point X,Y. You can get B

6=2(0)+b

B=6

The new equation is:

Y=2x+6

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Using the addition rule of the Sine function and the Cosine function, we obtain $\cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}$ .

<h3>What are the formulas for (sin x + sin y) and (cos x + cos y)?</h3>

The formula for the addition of two Sine functions ( $\sin x+\sin y$ ) is $\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$ .
The formula for the addition of two Cosine functions ( $\cos x+\cos y$ ) is $\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$ .

Given that

$\sin x + \sin y = a\\\cos x + \cos y = b$

Then using the above formulas, we get:

$2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=a$ (1)

$2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=b$ (2)

Dividing the equation (1) by (2), we get:

$\dfrac{\sin\dfrac{x+y}{2}}{\cos\frac{x-y}{2}}=\dfrac{a}{b}\\\Longrightarrow \tan\dfrac{x+y}{2}=\dfrac{a}{b}$ (3)

Now, we know that $\cos\theta=\dfrac{1}{\sqrt{1+\tan^2\theta}}$ .

Thus, using the above formula, we get from (3):

$\cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\tan^2\dfrac{x+y}{2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{1}{\sqrt{1+\dfrac{a^2}{b^2}}}\\\Longrightarrow \cos\dfrac{x+y}{2}=\dfrac{b}{\sqrt{a^2+b^2}}$