Solved on Nov 07, 2023

Find the sine of 292 degrees as a function of a positive acute angle.

STEP 1

Assumptions1. We are asked to express $\sin292^{\circ}$ as a function of a positive acute angle. . We know that the sine function is periodic with a period of $360^{\circ}$ .
3. We also know that $\sin (180^{\circ} + \theta) = -\sin \theta$ and $\sin (360^{\circ} - \theta) = -\sin \theta$ .

STEP 2

First, we need to find an equivalent angle to $292^{\circ}$ that lies within the first quadrant (0 to90 degrees). We can do this by subtracting multiples of $180^{\circ}$ or $360^{\circ}$ from $292^{\circ}$ until we get an angle in the first quadrant.

STEP 3

Subtract $180^{\circ}$ from $292^{\circ}$ .
$\theta =292^{\circ} -180^{\circ}$

STEP 4

Calculate the value of $\theta$ .
$\theta =292^{\circ} -180^{\circ} =112^{\circ}$

STEP 5

The angle $112^{\circ}$ is still not in the first quadrant. So, subtract $90^{\circ}$ from $112^{\circ}$ .
$\theta =112^{\circ} -90^{\circ}$

STEP 6

Calculate the value of $\theta$ .
$\theta =112^{\circ} -90^{\circ} =22^{\circ}$

STEP 7

Now, we have an acute angle $\theta =22^{\circ}$ . But we need to consider the negative sign because we have used the identity $\sin (180^{\circ} + \theta) = -\sin \theta$ .

STEP 8

Therefore, $\sin292^{\circ}$ can be expressed as a function of a positive acute angle as follows $\sin292^{\circ} = -\sin22^{\circ}$ So, $\sin292^{\circ}$ is equal to $-\sin22^{\circ}$ .

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Find the sine of 292 degrees as a function of a positive acute angle.

STEP 1

Assumptions1. We are asked to express $\sin292^{\circ}$ as a function of a positive acute angle. . We know that the sine function is periodic with a period of $360^{\circ}$ .
3. We also know that $\sin (180^{\circ} + \theta) = -\sin \theta$ and $\sin (360^{\circ} - \theta) = -\sin \theta$ .

STEP 2

First, we need to find an equivalent angle to $292^{\circ}$ that lies within the first quadrant (0 to90 degrees). We can do this by subtracting multiples of $180^{\circ}$ or $360^{\circ}$ from $292^{\circ}$ until we get an angle in the first quadrant.

STEP 3

Subtract $180^{\circ}$ from $292^{\circ}$ .
$\theta =292^{\circ} -180^{\circ}$

STEP 4

Calculate the value of $\theta$ .
$\theta =292^{\circ} -180^{\circ} =112^{\circ}$

STEP 5

The angle $112^{\circ}$ is still not in the first quadrant. So, subtract $90^{\circ}$ from $112^{\circ}$ .
$\theta =112^{\circ} -90^{\circ}$

STEP 6

Calculate the value of $\theta$ .
$\theta =112^{\circ} -90^{\circ} =22^{\circ}$

STEP 7

Now, we have an acute angle $\theta =22^{\circ}$ . But we need to consider the negative sign because we have used the identity $\sin (180^{\circ} + \theta) = -\sin \theta$ .

STEP 8

Therefore, $\sin292^{\circ}$ can be expressed as a function of a positive acute angle as follows $\sin292^{\circ} = -\sin22^{\circ}$ So, $\sin292^{\circ}$ is equal to $-\sin22^{\circ}$ .

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Find the sine of 292 degrees as a function of a positive acute angle.

STEP 1

STEP 2

First, we need to find an equivalent angle to 292∘292^{\circ}292∘ that lies within the first quadrant (0 to90 degrees). We can do this by subtracting multiples of 180∘180^{\circ}180∘ or 360∘360^{\circ}360∘ from 292∘292^{\circ}292∘ until we get an angle in the first quadrant.

STEP 3

Subtract 180∘180^{\circ}180∘ from 292∘292^{\circ}292∘.θ=292∘−180∘\theta =292^{\circ} -180^{\circ}θ=292∘−180∘

STEP 4

Calculate the value of θ\thetaθ.θ=292∘−180∘=112∘\theta =292^{\circ} -180^{\circ} =112^{\circ}θ=292∘−180∘=112∘

STEP 5

The angle 112∘112^{\circ}112∘ is still not in the first quadrant. So, subtract 90∘90^{\circ}90∘ from 112∘112^{\circ}112∘.θ=112∘−90∘\theta =112^{\circ} -90^{\circ}θ=112∘−90∘

STEP 6

Calculate the value of θ\thetaθ.θ=112∘−90∘=22∘\theta =112^{\circ} -90^{\circ} =22^{\circ}θ=112∘−90∘=22∘

STEP 7

Now, we have an acute angle θ=22∘\theta =22^{\circ}θ=22∘. But we need to consider the negative sign because we have used the identity sin⁡(180∘+θ)=−sin⁡θ\sin (180^{\circ} + \theta) = -\sin \thetasin(180∘+θ)=−sinθ.

STEP 8

Therefore, sin⁡292∘\sin292^{\circ}sin292∘ can be expressed as a function of a positive acute angle as followssin⁡292∘=−sin⁡22∘\sin292^{\circ} = -\sin22^{\circ}sin292∘=−sin22∘So, sin⁡292∘\sin292^{\circ}sin292∘ is equal to −sin⁡22∘-\sin22^{\circ}−sin22∘.

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First, we need to find an equivalent angle to $292^{\circ}$ that lies within the first quadrant (0 to90 degrees). We can do this by subtracting multiples of $180^{\circ}$ or $360^{\circ}$ from $292^{\circ}$ until we get an angle in the first quadrant.

Subtract $180^{\circ}$ from $292^{\circ}$ .
$\theta =292^{\circ} -180^{\circ}$

Calculate the value of $\theta$ .
$\theta =292^{\circ} -180^{\circ} =112^{\circ}$

The angle $112^{\circ}$ is still not in the first quadrant. So, subtract $90^{\circ}$ from $112^{\circ}$ .
$\theta =112^{\circ} -90^{\circ}$

Calculate the value of $\theta$ .
$\theta =112^{\circ} -90^{\circ} =22^{\circ}$

Now, we have an acute angle $\theta =22^{\circ}$ . But we need to consider the negative sign because we have used the identity $\sin (180^{\circ} + \theta) = -\sin \theta$ .

Therefore, $\sin292^{\circ}$ can be expressed as a function of a positive acute angle as follows $\sin292^{\circ} = -\sin22^{\circ}$ So, $\sin292^{\circ}$ is equal to $-\sin22^{\circ}$ .