Question

Sin(2/3 π - 2arctg 4/3)

Answer 1

$sin( \dfrac{2 \pi }{3} -2arctg \dfrac{4}{3})$

Пусть  $arctg \dfrac{4}{3}=x$
$sin( \dfrac{2 \pi }{3}-2x)=sin \dfrac{2 \pi }{3}cos2x-cos \dfrac{2 \pi }{3}sin2x= \dfrac{ \sqrt{3}cos2x+sin2x }{2}= \\ = \dfrac{ \sqrt{3}-2 \sqrt{3}sin^2x+2sinxcosx }{2}= \dfrac{ \sqrt{3} }{2}- \sqrt{3}sin^2x+sinxcosx \\ \\ tgx= \dfrac{4}{3} \Rightarrow cosx= \sqrt{ \dfrac{1}{1+tg^2x} }= \sqrt{ \dfrac{1}{1+ \frac{16}{9} } } = \dfrac{3}{5}$
$sinx= \sqrt{1-cos^2x}= \sqrt{1- \dfrac{9}{25} }= \dfrac{4}{5} \\ \\ \dfrac{ \sqrt{3} }{2}- \sqrt{3}sin^2x+sinxcosx= \dfrac{ \sqrt{3} }{2}- \sqrt{3}\cdot \dfrac{16}{25}+ \dfrac{4}{5}\cdot \dfrac{3}{5} = \dfrac{12}{25}- \dfrac{7 \sqrt{3} }{50}$

Ответ: $\dfrac{12}{25}- \dfrac{7 \sqrt{3} }{50}$

Answer 2

To find the value of the expression sin(2/3 π - 2arctg 4/3), we can start by using trigonometric identities to simplify it.

1. We know that arctan(x) is the inverse function of tangent, so we can write:

arctan(4/3) = θ

2. Now, we can express the angle (2/3 π - 2arctg 4/3) in terms of θ:

2/3 π - 2θ

3. We can use the following trigonometric identity for sine of the difference of two angles:

sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

4. Now, let's plug in our values:

sin(2/3 π - 2arctg 4/3) = sin(2/3 π)cos(2θ) - cos(2/3 π)sin(2θ)

5. Since sin(2/3 π) = sin(π/3) = sqrt(3)/2 and cos(2/3 π) = cos(π/3) = 1/2:

sin(2/3 π - 2arctg 4/3) = (sqrt(3)/2)cos(2θ) - (1/2)sin(2θ)

6. We also need to find cos(2θ) and sin(2θ). For that, we can use the double-angle trigonometric identities:

cos(2θ) = cos^2(θ) - sin^2(θ) sin(2θ) = 2sin(θ)cos(θ)

7. We already have the value of θ from step 1:

θ = arctan(4/3)

8. We can use the right triangle trigonometry to find sin(θ) and cos(θ):

Let's consider a right triangle with sides a = 4 and b = 3. The hypotenuse c can be found using the Pythagorean theorem: c^2 = a^2 + b^2. Thus, c^2 = 4^2 + 3^2 = 16 + 9 = 25. Taking the square root of both sides, c = √25 = 5.

Now, sin(θ) = b/c = 3/5, and cos(θ) = a/c = 4/5.

9. Now, we can find cos(2θ) and sin(2θ):

cos(2θ) = cos^2(θ) - sin^2(θ) = (4/5)^2 - (3/5)^2 = 16/25 - 9/25 = 7/25 sin(2θ) = 2sin(θ)cos(θ) = 2 * (3/5) * (4/5) = 24/25

10. Now, we can substitute these values back into our expression:

sin(2/3 π - 2arctg 4/3) = (sqrt(3)/2) * (7/25) - (1/2) * (24/25)

11. Now, calculate the expression:

sin(2/3 π - 2arctg 4/3) ≈ (sqrt(3)/2) * (7/25) - (1/2) * (24/25) ≈ (7sqrt(3)/50) - (12/50) ≈ (7sqrt(3) - 12) / 50

So, the approximate value of sin(2/3 π - 2arctg 4/3) is (7sqrt(3) - 12) / 50.