Sin^2x+cos^2 2x+sin^2 3x=3/2

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вопрос

The given expression is a trigonometric identity that can be proved using the Pythagorean trigonometric identity, which states that sin^2(x) + cos^2(x) = 1 for any angle x.

Starting with the given expression:
sin^2(x) + cos^2(2x) + sin^2(3x)

Using the double-angle identity for cosine, we can rewrite cos(2x) as 2cos^2(x) - 1:
sin^2(x) + 2cos^2(x) - 1 + sin^2(3x)

Now, we can use the triple-angle identity for sine to rewrite sin(3x) as 3sin(x) - 4sin^3(x):
sin^2(x) + 2cos^2(x) - 1 + (3sin(x) - 4sin^3(x))^2

Expanding the square term:
sin^2(x) + 2cos^2(x) - 1 + 9sin^2(x) - 24sin^4(x) + 16sin^6(x)

Simplifying further:
-24sin^4(x) + 16sin^6(x) + 10sin^2(x) + 2cos^2(x) - 1

Using the Pythagorean trigonometric identity sin^2(x) + cos^2(x) = 1, we know that cos^2(x) = 1 - sin^2(x). Substituting this into the expression:
-24sin^4(x) + 16sin^6(x) + 10sin^2(x) + 2(1 - sin^2(x)) - 1

-24sin^4(x) + 16sin^6(x) + 10sin^2(x) + 2 - 2sin^2(x) - 1

Combining like terms:
16sin^6(x) - 24sin^4(x) + 8sin^2(x) + 1

Now, simplify the expression to:
(4sin^2(x) - 1)^2

The final simplified expression is:
(4sin^2(x) - 1)^2

Therefore, sin^2(x) + cos^2(2x) + sin^2(3x) simplifies to (4sin^2(x) - 1)^2.