Jawab:
Penjelasan dengan langkah-langkah:
Identitas 1 :
[tex]\sin(x)+\cos(x) = \sqrt{2}\cdot \sin\left(x+\dfrac{\pi}{4}\right) = \sqrt{2}\cdot \cos\left(x-\dfrac{\pi}{4}\right)\\\\\boxed{(\sin(x)+\cos(x))^2 = 2\cdot \sin^2\left(x+\dfrac{\pi}{4}\right) = 2\cdot \cos^2\left(x-\dfrac{\pi}{4}\right)}[/tex]
Identitas 2 (lanjutan 1 dengan identitas pytagoras) :
[tex](\sin(x)+\cos(x))^2 = 2\cdot \left(1-\cos^2\left(x+\dfrac{\pi}{4}\right)\right) = 2\cdot \left(1-\sin^2\left(x-\dfrac{\pi}{4}\right)\right)\\\\\boxed{(\sin(x)+\cos(x))^2 = 2-2\cdot \cos^2\left(x+\dfrac{\pi}{4}\right) = 2-2\cdot \sin^2\left(x-\dfrac{\pi}{4}\right)}[/tex]
Identitas 3 :
[tex](\sin(x)+\cos(x))^2 = \sin^2(x)+\cos^2(x)+2\sin(x)\cos(x) \\\to \sin^2(x)+\cos^2(x) = 1, 2\sin(x)\cos(x) = \sin(2x)\\\\\boxed{(\sin(x)+\cos(x))^2 = 1+\sin(2x)}[/tex]
Identitas 4 (lanjutan 3):
[tex]\boxed{(\sin(x)+\cos(x))^2 = 1+\sin(2x) = \dfrac{1-\sin^2(2x)}{1-\sin(2x)} = \dfrac{\cos^2(2x)}{1-\sin(2x)}}[/tex]
atau :
[tex]\boxed{(\sin(x)+\cos(x))^2 = 1+\sin(2x) = \dfrac{1}{\sec^2(2x)-\tan(2x)\sec(2x)} = \dfrac{\cos(2x)}{\sec(2x)-\tan(2x)}}[/tex]
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