证明 (sin(t)+cos(t))^2=1+sin(2t)

解答

证明

(sin (t) + cos (t))^{2} = 1 + sin (2 t)

真

求解步骤

(sin (t) + cos (t))^{2} = 1 + sin (2 t)

调整右侧

1 + sin (2 t)

使用毕达哥拉斯恒等式:

1 = cos^{2} (t) + sin^{2} (t)

= (cos^{2} (t) + sin^{2} (t)) + sin (2 t)

使用倍角公式:

sin (2 t) = 2 sin (t) cos (t)

= cos^{2} (t) + sin^{2} (t) + 2 sin (t) cos (t)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

sin^{2} (t) + 2 sin (t) cos (t) + cos^{2} (t) = (sin (t) + cos (t))^{2}

= (sin (t) + cos (t))^{2}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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