# who to find the value of cos 3degree

### Asked by | 11th Aug, 2012, 10:54: PM

Expert Answer:

### 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.
**sin 15 degrees:**
By the half-angle 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.
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

**sin 18 degrees:**

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:**

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.

**cos 3 degrees:**

### Answered by | 12th Aug, 2012, 09:06: AM

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