Question

In Exercises 35–38, find the exact value of the following under the given conditions: d. sin 2α sin α = -1/3, 𝝅 \u003c α \u003c 3𝝅/2, and cos β = -1/3, 𝝅 \u003c β \u003c 3𝝅/2.

Accepted Answer

Identify the given information: $$ \sin \alpha = -\frac{1}{3} $$ with $$ \pi \u003c \alpha \u003c \frac{3\pi}{2} $$, and $$ \cos \beta = -\frac{1}{3} $$ with $$ \pi \u003c \beta \u003c \frac{3\pi}{2} . $$Note that both angles are in the third quadrant where sine and cosine are negative. Recall the double-angle identity for sine: $$ \sin 2\alpha = 2 \sin \alpha \cos \alpha . To $$find $$ \sin 2\alpha $$, we need $$ \cos \alpha  as $$well as $$ \sin \alpha . $$Use the Pythagorean identity to find $$ \cos \alpha $$: $$ \cos^2 \alpha = 1 - \sin^2 \alpha . $$Substitute $$ \sin \alpha = -\frac{1}{3}  to $$get $$ \cos^2 \alpha = 1 - \left(-\frac{1}{3}ight)^2 = 1 - \frac{1}{9} = \frac{8}{9} . $$Determine the sign of $$ \cos \alpha $$ based on the quadrant. Since $$ \alpha  is in $$the third quadrant, $$ \cos \alpha  is $$negative, so $$ \cos \alpha = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} . $$Substitute $$ \sin \alpha $$ and $$ \cos \alpha $$ into the double-angle formula: $$ \sin 2\alpha = 2 	imes \left(-\frac{1}{3}ight) 	imes \left(-\frac{2\sqrt{2}}{3}ight) . $$Simplify this expression to find the exact value of $$ \sin 2\alpha .$$

In Exercises 35–38, find the exact value of the following under the given conditions:
d. sin 2α
sin α = -1/3, 𝝅 < α < 3𝝅/2, and cos β = -1/3, 𝝅 < β < 3𝝅/2.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Double-Angle Identity for Sine

Determining the Sign of Trigonometric Functions Based on Quadrants

Using the Pythagorean Identity to Find Missing Trigonometric Values

In Exercises 35–38, find the exact value of the following under the given conditions: d. sin 2α sin α = -1/3, 𝝅 < α < 3𝝅/2, and cos β = -1/3, 𝝅 < β < 3𝝅/2.