Question

A) cos ² 15° - sin ²15° б) 2tgп/12 / 1+tg²п/12 Обчисліть

Answer 1

Відповідь:

cos²15° - sin²15° = cos(2*15°) = cos30° = √3/2

2tgπ/12 / (1 + tg²π/12) = 2tan(π/12)/(1+tan²(π/12)) = 2(2-√3)/(4-2√3+3) = 2(2-√3)/7

Пояснення:

Answer 2

A) cos² 15° - sin² 15°:

First, we can use the trigonometric identity for the square of the cosine and sine of the sum of two angles:

cos² α - sin² α = cos(2α)

In this case, α = 15°, so:

cos² 15° - sin² 15° = cos(2 * 15°) = cos(30°)

Now, we need to find the cosine of 30 degrees. Since cos(30°) is a commonly known value:

cos(30°) = √3 / 2

So, cos² 15° - sin² 15° = (√3 / 2)² - (1/2)² = 3/4 - 1/4 = 2/4 = 1/2

B) 2tg(π/12) / (1 + tg²(π/12)):

First, let's find the value of tangent (tg) of π/12:

tg(π/12) = sin(π/12) / cos(π/12)

Now, we can use the half-angle identities for sine and cosine:

sin(π/12) = √[(1 - cos(π/6))/2] = √[(1 - √3/2)/2] cos(π/12) = √[(1 + cos(π/6))/2] = √[(1 + √3/2)/2]

Now, calculate sin(π/12) and cos(π/12):

sin(π/12) = √[(1 - √3/2)/2] cos(π/12) = √[(1 + √3/2)/2]

Now, calculate tg(π/12):

tg(π/12) = sin(π/12) / cos(π/12)

Now, let's substitute these values into the expression:

2tg(π/12) / (1 + tg²(π/12)) = 2[√(1 - √3/2)/2] / [1 + (√(1 - √3/2)/√(1 + √3/2))^2]

Now, simplify further:

= √(2(1 - √3/2)) / [1 + (1 - √3/2) / (1 + √3/2)]

Now, simplify the expression in the denominator:

= √(2(1 - √3/2)) / [(1 + √3/2) + (1 - √3/2)]

= √(2(1 - √3/2)) / 2

Now, divide both the numerator and denominator by 2:

= √(1 - √3/2)

So, 2tg(π/12) / (1 + tg²(π/12)) = √(1 - √3/2)