Question

10sin10α/28cos5α , если sin5α=0,7

Answer 1

(5*sin5a*cos5a)/14cos5a=5sin5x/7= 0,7*5/7=0,5

Answer 2

To find the value of the expression $\frac{{10 \sin(10\alpha)}}{{28 \cos(5\alpha)}}$ when $\sin(5\alpha) = 0.7$, we'll use the given value of $\sin(5\alpha)$ and apply trigonometric identities.

Let's use the double-angle formula for sine and cosine:

1. $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
2. $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$

We'll need these formulas to simplify the expression.

First, let's find $\sin(10\alpha)$:

$\sin(10\alpha) = \sin(2\cdot5\alpha) = 2\sin(5\alpha)\cos(5\alpha)$

Now, substitute $\sin(5\alpha) = 0.7$:

$\sin(10\alpha) = 2 \cdot 0.7 \cdot \cos(5\alpha)$

Next, we need to find $\cos(5\alpha)$:

Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we can find $\cos(\theta)$ when $\sin(\theta)$ is known:

$\cos^2(\theta) = 1 - \sin^2(\theta)$

$\cos(5\alpha) = \sqrt{1 - \sin^2(5\alpha)}$

Substitute $\sin(5\alpha) = 0.7$:

$\cos(5\alpha) = \sqrt{1 - 0.7^2}$

Now, calculate $\sin(10\alpha)$:

$\sin(10\alpha) = 2 \cdot 0.7 \cdot \sqrt{1 - 0.7^2}$

Finally, plug $\sin(10\alpha)$ and $\cos(5\alpha)$ back into the original expression:

$\frac{{10 \sin(10\alpha)}}{{28 \cos(5\alpha)}} = \frac{{10 \cdot 2 \cdot 0.7 \cdot \sqrt{1 - 0.7^2}}}{{28 \cdot \sqrt{1 - 0.7^2}}}$

Now, simplify the expression:

$\frac{{10 \cdot 2 \cdot 0.7 \cdot \sqrt{1 - 0.7^2}}}{{28 \cdot \sqrt{1 - 0.7^2}}} = \frac{{2 \cdot 0.7 \cdot \sqrt{1 - 0.7^2}}}{{\sqrt{1 - 0.7^2}}}$

$= \frac{{2 \cdot 0.7}}{{1}}$

$= 1.4$

So, the value of the expression is $1.4$ when $\sin(5\alpha) = 0.7$.