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3 years ago

5 and six help me please

Mathematics

2 answers:

melisa1 [442]3 years ago

6 0

3,6,9,12,15,18,21,24,27,30
6,12,18,24,30,36,42,48,54,60
Common multiples are: 6,12,18,24,30

mina [271]3 years ago

3 0

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
Common: 20, 40

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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

$sin(3a)=3sin(a)-4sin^3(a)$

$cos(3a)=4cos^3(a)-3cos(a)$

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,

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

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