Question

1. for all x such that 0 ≤ x ≤ 90, which of the following expressions is not equal to sin x°? f. -sin(-x°) g. sin(-x°) h. cos(90 - x)° j. cos(x - 90)° k. √(1 - (cos x°)²) 43. a 3 - inch - tall rectangular box with a square base is constructed to hold a circular pie that has a diameter of 8 inches. both are shown below. what is the volume, in cubic inches, of the smallest such box that can hold this pie?

Explanation:

Step1: Analyze Option F

Use the property $\sin(-\alpha)=-\sin\alpha$. So, $-\sin(-x^{\circ})=-(-\sin x^{\circ})=\sin x^{\circ}$.

Step2: Analyze Option G

By the property $\sin(-\alpha)=-\sin\alpha$, $\sin(-x^{\circ})=-\sin x^{\circ}$, which is not equal to $\sin x^{\circ}$.

Step3: Analyze Option H

Use the co - function identity $\cos(90 - \alpha)=\sin\alpha$. So, $\cos(90 - x)^{\circ}=\sin x^{\circ}$.

Step4: Analyze Option J

$\cos(x - 90)^{\circ}=\cos(-(90 - x)^{\circ})$. Since $\cos(-\theta)=\cos\theta$, and $\cos(90 - x)^{\circ}=\sin x^{\circ}$, then $\cos(x - 90)^{\circ}=\sin x^{\circ}$.

Step5: Analyze Option K

By the Pythagorean identity $\sin^{2}\alpha+\cos^{2}\alpha = 1$, so $\sin\alpha=\sqrt{1-\cos^{2}\alpha}$ for $0\leq\alpha\leq90$. Thus, $\sqrt{1 - (\cos x^{\circ})^{2}}=\sin x^{\circ}$.

Step6: Solve problem 43

The diameter of the circular pie is 8 inches. Since the base of the rectangular box is square and it must hold the circular pie, the side - length of the square base of the box $s$ is equal to the diameter of the pie, so $s = 8$ inches. The height of the box $h=3$ inches.

Step7: Calculate the volume of the box

The volume of a rectangular box $V=s\times s\times h$. Substitute $s = 8$ inches and $h = 3$ inches into the formula: $V=8\times8\times3=192$ cubic inches.

Answer:

1. G. $\sin(-x^{\circ})$
2. C. 192