Question

Tg 2a=4/3 cos 2a=1/sqrt((1+(tg^2 ) 2a) чему равен cos 2a sqrt-квадратный корень tg^2-тангенс 2a в квадрате

Answer 1

$tg2a=\frac{4}{3}\\\\cos2a=\pm \frac{1}{\sqrt{1+tg^22a}}=\pm \frac{1}{\sqrt{1+\frac{16}{9}}}=\pm \frac{1}{\sqrt{\frac{25}{9}}}=\pm \frac{3}{5}\\\\\\\star \; \; 1+tg^2x=\frac{1}{cos^2x}\; \; \to \; \; cos^2x=\frac{1}{1+tg^2x}\; ,\; \; cosx=\pm \frac{1}{\sqrt{1+tg^2x}}\; \; \star$

Знак cos2a выбирается в зависимости от четверти, в которой лежит

угол 2а. Если угол находится в 1 и 4 четвертях, то сos2a>0, выбираем

знак (+). Если угол во 2 и 3 четвертях, то cos2a<0. выбираем знак (-).

Answer 2

Let's work on this step by step. We are given two equations:

1. tg(2a) = 4/3
2. cos(2a) = 1/sqrt(1 + (tg^2(2a)))

And we are asked to find the value of cos(2a) * sqrt(tg^2(2a)).

First, we need to find tg^2(2a) from the first equation:

tg^2(2a) = (4/3)^2 = 16/9

Now, let's find cos(2a) using the second equation:

cos(2a) = 1/sqrt(1 + tg^2(2a)) = 1/sqrt(1 + 16/9)

To simplify further, find the common denominator:

cos(2a) = 1/sqrt((9 + 16)/9) = 1/sqrt(25/9)

Now, take the square root of 25/9:

cos(2a) = 1/(5/3)

To divide by a fraction, you invert the divisor and multiply:

cos(2a) = 1 * (3/5) = 3/5

Now, we need to find cos(2a) * sqrt(tg^2(2a)):

cos(2a) * sqrt(tg^2(2a)) = (3/5) * sqrt(16/9)

To simplify further, find the square root:

cos(2a) * sqrt(tg^2(2a)) = (3/5) * (4/3)

Now, multiply:

cos(2a) * sqrt(tg^2(2a)) = 12/15

Finally, reduce the fraction:

cos(2a) * sqrt(tg^2(2a)) = 4/5

So, cos(2a) * sqrt(tg^2(2a)) is equal to 4/5.