1-sin²(x/2-3π)-cos²(x/4)+sin²(x/4)=(1-sin²(x/2-3π))-(cos²(x/4)-sin²(x/4))=
=cos²(x/2-3π)-cos(2*(x/4))=(cos(3π-x/2))²-cos(x/2)=(-cos(x/2))²-cos(x/2)=cos²(x/2)-cos(x/2)
<h3>sin²(π - x) + cos(π/2 + x) = 0</h3><h3>sin²x - sinx = 0</h3><h3>sinx•(sinx - 1) = 0</h3><h3>1) sinx = 0 ⇔ x = πn, n ∈ Z</h3><h3>2) sinx = 1 ⇔ x = π/2 + 2πk, k ∈ Z</h3><h3><u><em>ОТВЕТ: πn, n ∈ Z ; π/2 + 2πk, k ∈ Z</em></u></h3><h3><u><em /></u></h3>
<span>(4a-5b)(4a+5b)=16a^2-25b^2
Формула разности квадратов</span>