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Andrej [43]
3 years ago
12

Prove this identity sin(2A)=2sinAcosA.

Mathematics
1 answer:
34kurt3 years ago
7 0
The solution for proving the identity is as follows:

sin(2A) = sin(A + A) 
As sin(a + b) = sinacosb + sinbcosa, 
<span>sin(A + A) = sinAcosA + sinAcosA 
</span>
<span>Therefore, sin(2A) = 2sinAcosA

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your inquiries and questions soon. Have a nice day ahead!
</span>

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3 years ago
Side AB of parallelogram ABCD has endpoints at (-1,6) and (6,9). What is the slope of the opposite side, CD?
Butoxors [25]

Answer:

3/7

Step-by-step explanation:

Opposite sides on a parallelogram are parallel, and parallel lines have the same slope, so once we find the slope of AB, we'll know the slope of CD. Point A is (-1,6) and point B is (6,9), so the slope of AB (and by extension, CD) is

\dfrac{9-6}{6-(-1)}=\dfrac{3}{7}

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3 years ago
Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor
Alex787 [66]

The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

the lengths of the sides of BRAD are:

\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}

The lengths of the sides of KELY are:

\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}

Each side of KELY is twice the length of the corresponding side on BRAD.  This makes the ratio of the sides 2:1 and the figures are similar.

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3 years ago
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about 6000π

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