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lisabon 2012 [21]
3 years ago
11

I need help badly again....

Mathematics
2 answers:
Shkiper50 [21]3 years ago
5 0
Hello!

First of all we need to find the area of the circle (A=\pir²). Our radius is one. We square it which leaves us with \pi. Therefore we subtract pi from the triangle, which is 4√3.

Therefore, our answer is C) 4√3-\pi square units.

I hope this helps!
taurus [48]3 years ago
4 0
The second answer is correct.

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3 years ago
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Show so that the expression of cos4A in term of SinA is 8Sin^4 - 8Sin^2A + 1​
Degger [83]

Answer:

Step-by-step explanation:

L

H

S

=

cos

4

x

=

2

cos

2

(

2

x

)

−

1

=

2

(

cos

(

2

x

)

)

2

−

1

=

2

(

2

cos

2

x

−

1

)

2

−

1

=

2

(

4

cos

4

x

−

4

cos

2

x

+

1

)

−

1

=

8

cos

4

x

−

8

cos

2

x

+

2

−

1

=

8

cos

4

x

−

8

cos

2

x

+

1

=

R

H

S

Again

L

H

S

=

cos

4

x

=

2

cos

2

(

2

x

)

−

1

=

2

(

1

−

2

sin

2

x

)

)

2

−

1

=

2

(

1

−

4

sin

2

x

+

4

sin

4

x

)

−

1

=

2

−

8

sin

2

x

+

8

sin

4

x

−

1

=

8

sin

4

x

−

8

sin

2

x

+

1

=

R

H

S

sin

2

x

+

cos

2

x

=

1

cos

2

x

=

1

−

sin

2

x

substitute in the equation as follows

8

cos

4

x

−

8

cos

2

x

+

1

=

8

cos

2

x

(

cos

2

x

−

1

)

+

1

=

8

(

1

−

sin

2

x

)

(

1

−

sin

2

x

−

1

)

+

1

=

8

(

1

−

sin

2

x

)

(

−

sin

2

x

)

+

1

=

8

sin

4

x

−

8

sin

2

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