Question

Найти cos(a)^8-sin(a)^8, если cos(a)=1/3

Answer 1

(cos a)^8 - (sin a)^8 = [(cos a)^4 - (sin a)^4]*[(cos a)^4 + (sin a)^4] =
= (cos^2 a - sin^2 a)(cos^2 a + sin^2 a)*(cos^4 a + sin^4 a) =
= (cos 2a)*1*(cos^4 a + 2sin^2 a*cos^2 a + sin^4 a - 2sin^2 a*cos^2 a) =
= cos 2a *[ (cos^2 a + sin^2 a)^2 - 0,5*4sin^2 a*cos^2 a ] =
= cos 2a *(1^2 - 1/2*(sin 2a)^2) = cos 2a *(1 - 1/2*sin^2 (2a)) =
= cos 2a - 1/2*cos 2a*sin^2 (2a)
(cos a)^8 - (sin a)^8 = cos 2a - 1/2*cos 2a*sin^2 (2a)
Теперь подставляем. Так как cos a = 1/3, то:
cos 2a = 2cos^2 a - 1 = 2*1/9 - 1 = -7/9
sin^2 (2a) = 1 - cos^2 (2a) = 1 - 49/81 = 32/81
(cos a)^8 - (sin a)^8 = -7/9 - 1/2*(-7/9)*32/81 = -7/9 + 16*7/(9*81) =
= (-7*81+16*7)/729 = -455/729

Answer 2

To find the value of cos(a)^8 - sin(a)^8 when cos(a) = 1/3, we first need to determine the value of sin(a) using the Pythagorean identity for trigonometric functions:

$\cos^2(a) + \sin^2(a) = 1$

Since we know that $\cos(a) = \frac{1}{3}$, we can solve for $\sin(a)$:

$\sin^2(a) = 1 - \cos^2(a)$ $\sin^2(a) = 1 - \left(\frac{1}{3}\right)^2$ $\sin^2(a) = 1 - \frac{1}{9}$ $\sin^2(a) = \frac{8}{9}$

Taking the square root of both sides to get the positive value of $\sin(a)$:

$\sin(a) = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$

Now, we can substitute these values into the expression $\cos(a)^8 - \sin(a)^8$:

$\cos(a)^8 - \sin(a)^8 = \left(\frac{1}{3}\right)^8 - \left(\frac{2\sqrt{2}}{3}\right)^8$

Simplifying further:

$\cos(a)^8 - \sin(a)^8 = \frac{1}{3^8} - \frac{2^8 \cdot 2^4}{3^8}$

$\cos(a)^8 - \sin(a)^8 = \frac{1 - 2^4}{3^8}$

$\cos(a)^8 - \sin(a)^8 = \frac{1 - 16}{3^8}$

$\cos(a)^8 - \sin(a)^8 = \frac{-15}{3^8}$

$\cos(a)^8 - \sin(a)^8 = \frac{-15}{6561}$

So, when $\cos(a) = \frac{1}{3}$, the value of $\cos(a)^8 - \sin(a)^8$ is $\frac{-15}{6561}$.