Question

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

Answer 1

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

What are the formulas for (sin x + sin y) and (cos x + cos y)?

• 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):

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

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

To know more about Sine and Cosine functions, refer: brainly.com/question/27728553

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