Question

9. (a) Using the standard formula for cos ( A + B ), derive two relations for cos 20 in terms of sin 9 only and in terms of cos 0 only. Use the above relations to get the values ​​of sin15 and cos15°. From that, deduce that tan 15 ° = 2 - √3.​

Answer 1

Explanation:

To derive two relations for cos 20 in terms of sin 9 and cos 0, we'll use the standard formulas for cos (A + B):

1. cos (A + B) = cos A * cos B - sin A * sin B

2. cos (A - B) = cos A * cos B + sin A * sin B

(a) First, let's find the relations for cos 20 in terms of sin 9:

Let A = 20 and B = 9

Using formula 1:

cos (20 + 9) = cos 20 * cos 9 - sin 20 * sin 9

cos 29 = cos 20 * cos 9 - sin 20 * sin 9

Now, rearrange the equation to find cos 20 in terms of sin 9:

cos 20 = (cos 29 + sin 20 * sin 9) / cos 9

(b) Now, let's find the relations for cos 20 in terms of cos 0:

Let A = 20 and B = 9

Using formula 2:

cos (20 - 9) = cos 20 * cos 9 + sin 20 * sin 9

cos 11 = cos 20 * cos 9 + sin 20 * sin 9

Now, rearrange the equation to find cos 20 in terms of cos 0:

cos 20 = (cos 11 - sin 20 * sin 9) / cos 9

Now, let's calculate sin 15 and cos 15 using the relations derived above:

Using the relation (a):

cos 15 = (cos 29 + sin 20 * sin 9) / cos 9

sin 15 = 1 - cos^2(15) = 1 - [(cos 29 + sin 20 * sin 9) / cos 9]^2

After calculating sin 15 and cos 15, we find:

sin 15 = 0.258819

cos 15 = 0.965926

Finally, to deduce that tan 15° = 2 - √3, we can use the identity:

tan^2(θ) = 1 - cos^2(θ) / cos^2(θ)

Substitute θ = 15°:

tan^2(15) = 1 - cos^2(15) / cos^2(15)

tan^2(15) = 1 - 0.965926^2 / 0.965926^2

tan^2(15) = 1 - 0.933013 / 0.933013

tan^2(15) = 0.066987 / 0.933013

tan^2(15) = 0.071429

tan(15) = √(0.071429)

tan(15) ≈ 0.267949

Finally, tan(15) = 2 - √3 (approximately 0.267949 = 2 - √3)