Question

Tan=4, sin-? ............................​

Answer 1

Ответ:

± (4√17)/17

Пошаговое объяснение:

1) 1 + tg² α = 1/cos²α

1 + 4² = 1/cos²α

17 = 1/cos²α

cos²α = 1/17.

2) sinα = ± √(1 - cos²α) = ± √(1 - 1/17) = ± √(16/17) = ± (4√17)/17 ≈ ± 0,97

Ответ: ± (4√17)/17 ≈ ± 0,97

Answer 2

To find the sine of an angle when the tangent is given, you can use the following trigonometric identity:

$\sin^2(x) + \cos^2(x) = 1$

Since $\tan(x) = 4$, you can first find the value of $\cos(x)$ using the fact that $\tan(x) = \frac{\sin(x)}{\cos(x)}$:

$\frac{\sin(x)}{\cos(x)} = 4$

Now, let's square both sides:

$\frac{\sin^2(x)}{\cos^2(x)} = 4^2$

Using the identity $\sin^2(x) + \cos^2(x) = 1$, we can substitute:

$\frac{1 - \cos^2(x)}{\cos^2(x)} = 16$

Now, solve for $\cos(x)$:

$1 - \cos^2(x) = 16\cos^2(x)$

$1 = 17\cos^2(x)$

$\cos^2(x) = \frac{1}{17}$

Now, take the square root of both sides, but since we're dealing with an angle, consider both positive and negative square roots:

$\cos(x) = \pm \sqrt{\frac{1}{17}}$

Now that you have $\cos(x)$, you can find $\sin(x)$ using the fact that $\sin(x) = \sqrt{1 - \cos^2(x)}$:

$\sin(x) = \sqrt{1 - \left(\pm \sqrt{\frac{1}{17}}\right)^2}$

Since the sine function is always positive, you can take the positive square root:

$\sin(x) = \sqrt{1 - \frac{1}{17}}$

Now, simplify:

$\sin(x) = \sqrt{\frac{16}{17}}$

$\sin(x) = \frac{4}{\sqrt{17}}$

To rationalize the denominator, multiply both the numerator and denominator by $\sqrt{17}$:

$\sin(x) = \frac{4\sqrt{17}}{17}$

So, $\sin(x) = \frac{4\sqrt{17}}{17}$ when $\tan(x) = 4$.