If sin theta = sqrt(2/2), which could not be the value of theta?
-225 degrees
-45 degrees
-135 degrees
-405 degrees

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BlueSky06 BlueSky06 · Answer 1 · 2017-05-31T15:02:01+02:00

The solution would be like this for this specific problem:

sin(θ°) = √(2)/2

θ° = 360°n + sin⁻¹(√(2)/2) and θ° = 360°n + 180° − sin⁻¹(√(2)/2)
θ° = 360°n + 45° and θ° = 360°n + 135° where n∈ℤ

360°*0 + 45° = 45°
360°*0 + 135° = 135°
360°*1 + 45° = 405°

sin(225°) = -√(2)/2

225 has an angle where sin theta= -(sqrt2)/2 therefore, the value of theta cannot be 225 degrees.

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SerenaBochenek SerenaBochenek · Answer 2 · 2019-05-27T11:15:09+02:00

Answer:

225° is not possible

Step-by-step explanation:

Given that

[tex]\sin \theta=\frac{\sqrt{2}}{2}[/tex]

we have to choose the option which could not be the value of theta.

[tex]\sin \theta=\frac{\sqrt{2}}{2}[/tex]

[tex]\sin \theta=\frac{1}{\sqrt2}=\sin45^{\circ}[/tex]

[tex]\theta=45^{\circ}[/tex]

As sine is positive in second and fourth quadrant.

⇒ [tex]\sin 45=\sin(180-45)=\sin135[/tex]

Also, [tex]\sin 45=\sin(360+45)=\sin405[/tex]

[tex]\text{Hence, the value of }\theta \text{which are possible are } 45^{\circ}, 135^{\circ}, 405^{\circ}[/tex]

Therefore 225° is not possible