Question

Решите cos 165° · cos 285° Тоже решите sin 75° + sin 15°

Answer 1

1) cos(285*)*cos(165*) = (1/2)*( cos(285*+165*) + cos(285*-165*) ) == (1/2)*( cos(450*) + cos(120*) ) = (1/2)*( cos(360*+90*) + cos(180*-60*) ) = (1/2)*(cos(90*) - cos(60*) ) = (1/2)*( 0 - (1/2)) = -1/4 = -0,25

2)sin75+sin 15 = 2sin (75+15)/2 * cos (75-15)/2 = 2sin 45*cos30

Answer 2

$\mathtt{cos165аcos285а=\frac{cos(165а-285а)+cos(165а+285а)}{2}=}\\\mathtt{\frac{cos120а+cos450а}{2}=\frac{cos(180а-60а)+cos(360а+90а)}{2}=-\frac{cos60а}{2}=-\frac{1}{4}}$

$\mathtt{sin75а+sin15а=sin15а+cos15а=\sqrt{2}sin60а=\frac{\sqrt{6}}{2}=\sqrt{1,5}}$

Answer 3

To solve these trigonometric expressions, we'll use the following trigonometric identities:

1. Product-to-Sum Identity: cos(A) · cos(B) = (1/2) * [cos(A - B) + cos(A + B)]
2. Sum-to-Product Identity: sin(A) + sin(B) = 2 * sin((A + B)/2) * cos((A - B)/2)

Let's start with the first expression:

1. cos 165° · cos 285°

Using the product-to-sum identity, we can rewrite the expression as:

cos 165° · cos 285° = (1/2) * [cos(165° - 285°) + cos(165° + 285°)]

Now, simplify the angles inside the cosines:

cos(165° - 285°) = cos(-120°) = cos(240°) [cosine function is even] cos(165° + 285°) = cos(450°) = cos(360° + 90°) = cos(90°) = 0 [cosine function is periodic]

The expression becomes:

(1/2) * [cos(240°) + 0] = (1/2) * cos(240°)

Now, we need to find the cosine of 240°. Recall that the cosine function has a period of 360°, and the cosine function is negative in the second quadrant (180° to 270°). Therefore, cos(240°) = -cos(240° - 180°) = -cos(60°).

cos(60°) = 1/2

So, the final result is:

cos 165° · cos 285° = (1/2) * (1/2) = 1/4

Now, let's solve the second expression:

2. sin 75° + sin 15°

Using the sum-to-product identity, we can rewrite the expression as:

sin 75° + sin 15° = 2 * sin((75° + 15°)/2) * cos((75° - 15°)/2)

Simplify the angles inside the sines and cosines:

sin(75° + 15°) = sin(90°) = 1 [sine function reaches maximum at 90°] cos(75° - 15°) = cos(60°) = 1/2

The expression becomes:

2 * sin(90°/2) * cos(60°/2) = 2 * sin(45°) * cos(30°)

Now, let's find the values of sin(45°) and cos(30°):

sin(45°) = √2 / 2 cos(30°) = √3 / 2

So, the final result is:

sin 75° + sin 15° = 2 * (√2 / 2) * (√3 / 2) = √6

Therefore, the solutions are:

1. cos 165° · cos 285° = 1/4
2. sin 75° + sin 15° = √6