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Readme [11.4K]
3 years ago
14

Can anyone please show me how to find sin 18 in fraction form? plzzzzzzz $_$​

Mathematics
1 answer:
Inessa [10]3 years ago
7 0

Answer:

(-1 + √5) / 4

Step-by-step explanation:

18° is 1/5 of 90°. So if we say x = 18:

5x = 90

You probably won't find an identity for sin(5x) in your textbook.  But you will find double angle and triple angle formulas.  So if we split 5x into two terms:

2x + 3x = 90

Rearrange:

2x = 90 − 3x

Take sine of both sides:

sin(2x) = sin(90 − 3x)

Use phase shift identity:

sin(2x) = cos(3x)

Apply double and triple angle formulas:

2 sin x cos x = 4 cos³ x − 3 cos x

Simplify:

0 = 4 cos³ x − 3 cos x − 2 sin x cos x

0 = cos x (4 cos² x − 3 − 2 sin x)

We know cos 18° isn't 0, so we can divide it out:

0 = 4 cos² x − 3 − 2 sin x

Using Pythagorean identity and simplifying:

0 = 4 (1 − sin² x) − 3 − 2 sin x

0 = 4 − 4 sin² x − 3 − 2 sin x

0 = 1 − 4 sin² x − 2 sin x

0 = 4 sin² x + 2 sin x − 1

Solve with quadratic formula (or you can complete the square):

sin x = [ -2 ± √(2² − 4(4)(-1)) ] / 2(4)

sin x = [ -2 ± √(4 + 16) ] / 8

sin x = (-2 ± √20) / 8

sin x = (-2 ± 2√5) / 8

sin x = (-1 ± √5) / 4

18° is in the first quadrant, so we know sin x > 0.  Therefore:

sin x =  (-1 + √5) / 4

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1) 3 + 2i + 2 - 5i = 3 +2 + 2i - 5i

                          = 5 + (2-5)i

                          = 5 + (-3)i

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7) (3 - 2i)² + (3 +2i) = 3² - 2*3*2i + (2i)² + 3 + 2i     {(a - b)² = a² - 2ab +b²}

= 9 -12i + 4i² + 3 + 2i

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= 9 +3 - 4 - 12i +2i

= 8 - 10i

9) (a +b)^{4} - (a - b)^{4} = 8ab(a^{2} +b^{2})\\\\\\Here, \ a = 1 \ and \ b = i\\\\\\(1+\sqrt{-1})^{4}-(1-\sqrt{-1})^{4}=(1+\sqrt{i^{2}})^{4}-(1-\sqrt{i^{2}})^{4}\\\\=(1+i)^{4}-(1-i)^{4}\\\\=8*1*i(1^{2}+i^{2})\\\\=8i*(1 -1)\\\\=8i*0\\\\=0

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