Question

Вычислить sin(arccos(1/4))

Answer 1

Ответ:

Объяснение:

По формуле: $sin(arccos(A)) = \sqrt{1-x^{2} }$

$sin(arccos(1/4)=\sqrt{1-\frac{1}{16} } =\sqrt{\frac{16-1}{16} } =\frac{\sqrt{15} }{4}$

Answer 2

$sin(\underbrace {arccos \dfrac{1}{4}}_{\alpha })=sin\alpha\\\\\alpha =arccos \dfrac{1}{4}\ \ \ \Rightarrow \ \ \ cos\alpha =\dfrac{1}{4}>0\ \ \Rightarrow \ \ \ 0<\alpha <\dfrac{\pi}{2}\\\\\\sin^2\alpha +cos^2\alpha =1\ \ \to \ \ \ sin^2\alpha =1-cos^2\alpha \ \ ,\ \ sin\alpha =\pm \sqrt{1-cos^2\alpha }\ \ ,\\\\0<\alpha <\dfrac{\pi}{2}\ \ \Rightarrow \ \ sin\alpha =+\sqrt{1-cos^2\alpha }=\sqrt{1-\dfrac{1}{16}}=\dfrac{\sqrt{15}}{4}$