Sin^2 40°+sin^2 50°+sin^2 230°+sin^2 320°

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вопрос

= sin^2 40° + sin^2 50° + sin^2 (230°+180°) + sin^2 (320°-360°)
= sin^2 40° + sin^2 50° + sin^2 50° + sin^2 320°
= sin^2 40° + 2(sin^2 50°) + sin^2 320°
Using the trigonometric identity sin^2 θ + cos^2 θ = 1, we have:
= (1 - cos^2 40°) + 2(1 - cos^2 50°) + (1 - cos^2 320°)
= 3 - cos^2 40° + 2 - 2cos^2 50° + 3 - cos^2 320°
= 8 - cos^2 40° - 2cos^2 50° - cos^2 320°

Now, we can use the fact that sin(x) = cos(90° - x) to simplify further.
= 8 - sin^2 50° - 2sin^2 40° - sin^2 40°
= 8 - 3sin^2 40° - sin^2 50°

Therefore, sin^2 40° + sin^2 50° + sin^2 230° + sin^2 320° = 8 - 3sin^2 40° - sin^2 50°.