Answer on Question #58925 – Math – Trigonometry
Question
1. Evaluate sin − 1 ( 2 2 ) \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) sin − 1 ( 2 2 )
Solution
If sin − 1 \sin^{-1} sin − 1
sin − 1 ( 2 2 ) = π 4 , that is, sin − 1 ( 2 2 ) = 45 ∘ . \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}, \text{ that is, } \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45{}^\circ. sin − 1 ( 2 2 ) = 4 π , that is, sin − 1 ( 2 2 ) = 45 ∘ .
If sin − 1 ( x ) = 1 sin ( x ) \sin^{-1}(x) = \frac{1}{\sin(x)} sin − 1 ( x ) = s i n ( x ) 1
sin − 1 ( 2 2 ) = 1 sin ( 2 2 ) ≈ 1 sin ( 0.707 ) ≈ 1 0.65 ≈ 1.539. \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{1}{\sin\left(\frac{\sqrt{2}}{2}\right)} \approx \frac{1}{\sin(0.707)} \approx \frac{1}{0.65} \approx 1.539. sin − 1 ( 2 2 ) = sin ( 2 2 ) 1 ≈ sin ( 0.707 ) 1 ≈ 0.65 1 ≈ 1.539.
Answer: π 4 \frac{\pi}{4} 4 π
Question
2. Is the value π 16 \frac{\pi}{16} 16 π 2 cos 2 ( 4 x ) − 1 = 0 2\cos^2(4x) - 1 = 0 2 cos 2 ( 4 x ) − 1 = 0
False
True
Solution
Let x = π 16 x = \frac{\pi}{16} x = 16 π
2 cos 2 ( 4 x ) − 1 = 2 cos 2 ( 4 ⋅ π 16 ) − 1 = 2 cos 2 ( π 4 ) − 1 = 2 ( 2 2 ) 2 − 1 = 2 ⋅ 2 4 − 1 = 1 − 1 = 0 ⇒ ⇒ x = π 16 is a solution for the equation 2 cos 2 ( 4 x ) − 1 = 0 , hence it is true . \begin{aligned}
2 \cos^2(4x) - 1 &= 2 \cos^2\left(4 \cdot \frac{\pi}{16}\right) - 1 \\
&= 2 \cos^2\left(\frac{\pi}{4}\right) - 1 \\
&= 2 \left(\frac{\sqrt{2}}{2}\right)^2 - 1 \\
&= 2 \cdot \frac{2}{4} - 1 \\
&= 1 - 1 = 0 \Rightarrow \\
&\Rightarrow x = \frac{\pi}{16} \text{ is a solution for the equation } 2 \cos^2(4x) - 1 = 0, \text{ hence it is true}.
\end{aligned} 2 cos 2 ( 4 x ) − 1 = 2 cos 2 ( 4 ⋅ 16 π ) − 1 = 2 cos 2 ( 4 π ) − 1 = 2 ( 2 2 ) 2 − 1 = 2 ⋅ 4 2 − 1 = 1 − 1 = 0 ⇒ ⇒ x = 16 π is a solution for the equation 2 cos 2 ( 4 x ) − 1 = 0 , hence it is true .
Answer: True.
Question
3. Is the value 3 π 2 \frac{3\pi}{2} 2 3 π 2 sin 2 ( x ) − sin ( x ) − 1 = 0 2\sin^2(x) - \sin(x) - 1 = 0 2 sin 2 ( x ) − sin ( x ) − 1 = 0
Solution
Let x = 3 π 2 x = \frac{3\pi}{2} x = 2 3 π 2 sin 2 ( x ) − sin ( x ) − 1 = 2 sin 2 ( 3 π 2 ) − sin ( 3 π 2 ) − 1 = 2\sin^2(x) - \sin(x) - 1 = 2\sin^2\left(\frac{3\pi}{2}\right) - \sin\left(\frac{3\pi}{2}\right) - 1 = 2 sin 2 ( x ) − sin ( x ) − 1 = 2 sin 2 ( 2 3 π ) − sin ( 2 3 π ) − 1 =
= 2 sin 2 ( π + π 2 ) − sin ( π + π 2 ) − 1 = 2 ( − 1 ) 2 − ( − 1 ) − 1 = 2 + 1 − 1 = 2 ≠ 0 ⇒ x = 3 π 2 is not a solution for the equation 2 sin 2 ( x ) − sin ( x ) − 1 = 0 , hence it is false . = 2 \sin^2\left(\pi + \frac{\pi}{2}\right) - \sin\left(\pi + \frac{\pi}{2}\right) - 1 = 2(-1)^2 - (-1) - 1 = 2 + 1 - 1 = 2 \neq 0 \Rightarrow x = \frac{3\pi}{2} \text{ is not a solution for the equation } 2 \sin^2(x) - \sin(x) - 1 = 0, \text{ hence it is false}. = 2 sin 2 ( π + 2 π ) − sin ( π + 2 π ) − 1 = 2 ( − 1 ) 2 − ( − 1 ) − 1 = 2 + 1 − 1 = 2 = 0 ⇒ x = 2 3 π is not a solution for the equation 2 sin 2 ( x ) − sin ( x ) − 1 = 0 , hence it is false .
Answer: False.
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#339914 on Dec 2023