Sin(2arccos 1/5)

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вопрос

Let's first find the value of arccos(1/5):

cos(arccos(1/5)) = 1/5

Since the cosine of an arccos function is equal to the value inside the arccos function, we know that cos(arccos(1/5)) = 1/5.

Now, we can calculate the value of sin(2arccos(1/5)) using the double angle identity:

sin(2arccos(1/5)) = 2sin(arccos(1/5))cos(arccos(1/5))

sin(arccos(1/5)) can be found using the Pythagorean identity:

sin^2(arccos(1/5)) + cos^2(arccos(1/5)) = 1

sin^2(arccos(1/5)) + (1/5)^2 = 1
sin^2(arccos(1/5)) + 1/25 = 1
sin^2(arccos(1/5)) = 24/25
sin(arccos(1/5)) = ±√(24/25)
sin(arccos(1/5)) = ±√24/√25
sin(arccos(1/5)) = ±√24/5

Now, we substitute these values back into the original equation:

sin(2arccos(1/5)) = 2(±√24/5)(1/5)
sin(2arccos(1/5)) = ±2√24/25

Therefore, sin(2arccos(1/5)) could be expressed as ±2√24/25.