Respuesta :
Answer:
[tex]\frac{24}{25}[/tex]
Step-by-step explanation:
Given
cosA = [tex]\frac{3}{5}[/tex] = [tex]\frac{adjacent}{hypotenuse}[/tex]
Then the opposite side using Pythagorean triple is 4, thus
sinA = [tex]\frac{4}{5}[/tex]
Using the trigonometric identity
sin2A = 2 sinAcosA, then
sin2A = 2 × [tex]\frac{4}{5}[/tex] × [tex]\frac{3}{5}[/tex]
= [tex]\frac{2(4)(3)}{5(5)}[/tex] = [tex]\frac{24}{25}[/tex]
Answer:
[tex]\sin(2A)=\frac{24}{25}[/tex]
Step-by-step explanation:
Double angle identity for sine is
[tex]\sin(2A)=2\sin(A)\cos(A)[/tex].
We know [tex]\cos(A)[/tex]. We now need to find [tex]\sin(A)[/tex] to find our answer.
Recall the following [tex]Pythagorean Identity[/tex]:
[tex]\sin^2(A)+\cos^2(A)=1[/tex].
Replace [tex]\cos(A)[/tex] with [tex]\frac{3}{5}[/tex]:
[tex]\sin^2(A)+(\frac{3}{5})^2=1[/tex]
[tex]\sin^2(A)+\frac{9}{25}=1[/tex]
Subtract [tex]\frac{9}{25}[/tex] on both sides:
[tex]\sin^2(A)=1-\frac{9}{25}[/tex]
[tex]\sin^2(A)=\frac{25}{25}-\frac{9}{25}[/tex]
[tex]\sin^2(A)=\frac{25-9}{25}[/tex]
[tex]\sin^2(A)=\frac{16}{25}[/tex]
Now to finally get the value of [tex]\sin(A)[/tex] take the square of both sides:
[tex]\sin(A)=\pm \sqrt{\frac{16}{25}}[/tex]
Since [tex]A[/tex] is in the interval [tex](0,90)[/tex], then [tex]\sin(A)>0[/tex].
[tex]\sin(A)=\sqrt{\frac{16}{25}}[/tex]
[tex]\sin(A)=\frac{\sqrt{16}}{\sqrt{25}}[/tex]
[tex]\sin(A)=\frac{4}{5}[/tex]
So let's finally find the numerical value of [tex]\sin(2A)[/tex].
[tex]\sin(2A)=2\sin(A)\cos(A)[/tex]
[tex]\sin(2A)=2\cdot \frac{4}{5} \cdot \frac{3}{5}[/tex]
[tex]\sin(2A)=\frac{2(4)(3)}{5(5)}[/tex]
[tex]\sin(2A)=\frac{24}{25}[/tex]
Side note:
If [tex]A[/tex] is between 0 and 90 degrees, then we are in the first quadrant.
If we are in the first quadrant, both [tex]x=\cos(A)[/tex] and [tex]y=\sin(A)[/tex] are positive.
