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Brums [2.3K]
3 years ago
14

How do I solve this problem?

Mathematics
1 answer:
Leokris [45]3 years ago
8 0

Answer:

21, 18, 15

Step-by-step explanation:

First term = a1

Second term = a2

Third term = a3

a1 = 21

a2

= a + d

= 21 - 3

= 18

a3

= a + 2d

= 21 + 2 ( - 3 )

= 21 - 6

= 15

The first 3 terms are 21, 18, 15.

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Answer:

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

Step-by-step explanation:

we are given

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

we can simplify left side and make it equal to right side

we can use trig identity

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now, we can plug values

\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}

now, we can simplify

\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}

\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}

\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}

now, we can factor it

\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}

\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}

we can use trig identity

sin^2(a)+cos^2(a)=1

\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}

we can cancel terms

=3-4(1-sin(a)cos(a))

now, we can simplify it further

=3-4+4sin(a)cos(a))

=-1+4sin(a)cos(a))

=4sin(a)cos(a)-1

=2\times 2sin(a)cos(a)-1

now, we can use trig identity

2sin(a)cos(a)=sin(2a)

we can replace it

=2sin(2a)-1

so,

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1


7 0
3 years ago
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Determine the coordinates of the vertices of the rectangle to compite the area of the rectangle using the distance formula.​
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Answer:

Step-by-step explanation:

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3 years ago
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7 0
4 years ago
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Answer:

A

Step-by-step explanation:

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5 0
3 years ago
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