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Express as a function of a DIFFERENT angle, 0^(@) <= theta < 360^(@). sin(309^(@)) sin(◻^(@))

Express as a function of a DIFFERENT angle, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} .\newlinesin⁡(309∘) \sin \left(309^{\circ}\right) \newlinesin⁡(□∘) \sin \left(\square^{\circ}\right)

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Full solution

Q. Express as a function of a DIFFERENT angle, 0∘≤θ<360∘ 0^{\circ} \leq \theta<360^{\circ} .\newlinesin⁡(309∘) \sin \left(309^{\circ}\right) \newlinesin⁡(□∘) \sin \left(\square^{\circ}\right)
  1. Understand the problem: Understand the problem.\newlineWe need to express sin⁡(309∘)\sin(309^\circ) in terms of a different angle that is within the range of 0∘0^\circ to 360∘360^\circ. This typically involves using known trigonometric identities or properties to rewrite the sine of an angle in terms of another angle.
  2. Use trigonometric identities: Use the fact that sin⁡(θ)=sin⁡(180°−θ)\sin(\theta) = \sin(180° - \theta) or sin⁡(360°−θ)\sin(360° - \theta) to find an equivalent expression.\newlineSince 309°309° is more than 180°180°, we can use the identity sin⁡(θ)=sin⁡(360°−θ)\sin(\theta) = \sin(360° - \theta) to find a different angle that gives the same sine value.
  3. Calculate equivalent angle: Calculate the equivalent angle using the identity.\newlinesin⁡(309°)=sin⁡(360°−309°)\sin(309°) = \sin(360° - 309°)\newlinesin⁡(309°)=sin⁡(51°)\sin(309°) = \sin(51°)
  4. Verify new angle: Verify that the new angle is within the specified range.\newline51°51° is indeed within the range of 0°0° to 360°360°, so our expression is valid.

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