Cos7x cos8x + sin7x sin8x = корень из 3/2

Предыдущий
вопрос Следующий
вопрос

To prove this identity, we start by applying the angle sum identity for cosine:

cos(A + B) = cosA cosB - sinA sinB

Here, A = 7x and B = 8x:

cos(7x + 8x) = cos7x cos8x - sin7x sin8
cos15x = cos7x cos8x - sin7x sin8x

Now, let's apply the angle sum identity for sine:

sin(A + B) = sinA cosB + cosA sinB

Again, A = 7x and B = 8x:

sin(7x + 8x) = sin7x cos8x + cos7x sin8
sin15x = sin7x cos8x + cos7x sin8x

We know that cos(15x) = sqrt(3)/2 and sin(15x) = 1/2 (using 15 degrees as reference angles in the unit circle). Therefore:

cos7x cos8x - sin7x sin8x = sqrt(3)/
sin7x cos8x + cos7x sin8x = 1/2

We now have a system of two equations:

1) cos7x cos8x - sin7x sin8x = sqrt(3)/
2) sin7x cos8x + cos7x sin8x = 1/2

We can now square both equations and add them together to simplify:

(cos7x cos8x)^2 - 2(cos7x cos8x)(sin7x sin8x) + (sin7x sin8x)^2 + (sin7x cos8x)^2 + 2(sin7x cos8x)(cos7x sin8x) + (cos7x sin8x)^2 = (sqrt(3))^2/2^2 + (1)^2/2^
(cos7x cos8x)^2 - 2(cos7x cos8x)(sin7x sin8x) + (sin7x sin8x)^2 + (sin7x cos8x)^2 + 2(sin7x cos8x)(cos7x sin8x) + (cos7x sin8x)^2 = 3/4 + 1/
(cos7x cos8x - sin7x sin8x)^2 + (sin7x cos8x + cos7x sin8x)^2 = 1

Hence, we have proved that cos(7x)cos(8x) + sin(7x)sin(8x) = sqrt(3)/2.