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 who to find the value of cos 3degree
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First, find sin (18) and sin (15). Then use these to find cos (18) and cos (15); and then finally you can easily find sin (3).
sin 18 degrees:
sin (72) = 2 sin (36) cos (36)
By the double angle formula for sines again, 2 sin (36) = 4 sin (18) cos (18); and by the double angle formula for cosines, cos (36) = 1  2 sin^2 (18). Thus, we have
sin (72) = (4 sin (18) cos (18)) (1  2 sin^2 (18))
cos (18) = 4 sin (18) cos (18) (1  2 sin^2 (18))
Divide both sides by cos (18):
1 = 4 sin (18) (1  2 sin^2 (18))
Let x = sin 18
1 = 4x(1  2x^2)
1 = 4x  8x^3
8x^3  4x + 1 = 0
(2x  1)(4x^2 + 2x  1) = 0
So we get x = 1/2 as one solution, and the quadratic formula gives the other solutions as x = (1 + ?5) / 4 and x = (1  ?5) / 4.
Now, we know that sin 18 has to be one of these values.
sin (18) can't be 1/2, because sin (0) = 0, sin (30) = 1/2, and sin x is an increasing function between 0 and 30.
sin (18) can't be (1  ?5) / 4, because sin x is positive when x is between 0 and 180.
Therefore, sin (18) = (1 + ?5) / 4.
By the double angle formula for sines again, 2 sin (36) = 4 sin (18) cos (18); and by the double angle formula for cosines, cos (36) = 1  2 sin^2 (18). Thus, we have
sin (72) = (4 sin (18) cos (18)) (1  2 sin^2 (18))
cos (18) = 4 sin (18) cos (18) (1  2 sin^2 (18))
Divide both sides by cos (18):
1 = 4 sin (18) (1  2 sin^2 (18))
Let x = sin 18
1 = 4x(1  2x^2)
1 = 4x  8x^3
8x^3  4x + 1 = 0
(2x  1)(4x^2 + 2x  1) = 0
So we get x = 1/2 as one solution, and the quadratic formula gives the other solutions as x = (1 + ?5) / 4 and x = (1  ?5) / 4.
Now, we know that sin 18 has to be one of these values.
sin (18) can't be 1/2, because sin (0) = 0, sin (30) = 1/2, and sin x is an increasing function between 0 and 30.
sin (18) can't be (1  ?5) / 4, because sin x is positive when x is between 0 and 180.
Therefore, sin (18) = (1 + ?5) / 4.
sin 15 degrees:
By the halfangle formula for sines, we have
sin (30/2) = ?{(1  cos 30) / 2}
But cos (30) = ?3 / 2, so we have:
sin (15) = ?{(1  ?3 / 2) / 2}
sin (15) = ?((2/2  ?3 / 2) / 2}
sin (15) = ?{(2  ?3) / 4}
sin (15) = ?(2 ?3) / 2.
It turns out that ?(2 ?3) is the same as (?6  ?2) / 2. This is not obvious, but if you square (?6  ?2) / 2 and simplify, it turns out that you get 2 ?3. So the answer can be simplified a bit further:
sin (15) = (?6  ?2) / 4.
sin (30/2) = ?{(1  cos 30) / 2}
But cos (30) = ?3 / 2, so we have:
sin (15) = ?{(1  ?3 / 2) / 2}
sin (15) = ?((2/2  ?3 / 2) / 2}
sin (15) = ?{(2  ?3) / 4}
sin (15) = ?(2 ?3) / 2.
It turns out that ?(2 ?3) is the same as (?6  ?2) / 2. This is not obvious, but if you square (?6  ?2) / 2 and simplify, it turns out that you get 2 ?3. So the answer can be simplified a bit further:
sin (15) = (?6  ?2) / 4.
Now, use the values of sin (18) and sin (15) to find the values of cos (18) and cos (15). Make use of the identity sin^2 x + cos^2 x = 1.
You will get:
cos (18) = ?{(10 + 2?5)} / 4
cos (15) = (?6 + ?2) / 4
cos 3 degrees:
To find cos 3, use the identity
cos (A  B) = cos A cos B + sin A sin B
Put A = 18 and B = 15
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