Select the appropriate inequality symbol to make the number sentence true.

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

It’s not equal it’s the one on the left

Serggg [28]3 years ago

7 0

Answer:

Step-by-step explanation:

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Ainat [17]

Answer:

d.25 that is the answer of the question

5 0

3 years ago

Sinx + siny=a<br> cosx + cosy=b<br> Find cos(x+y/2)

romanna [79]

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):

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