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find an angle \\( \\theta \\) with \\( 0^{\\circ}<\\theta<360^{\\circ} …

Question

find an angle \\( \theta \\) with \\( 0^{\circ}<\theta<360^{\circ} \\) that has the same:
sine function value as \\( 190^{\circ} \\)
\\( \theta= \\) degrees
cosine function value as \\( 190^{\circ} \\)
\\( \theta= \\) degrees
question help: video

Explanation:

Step1: Find the angle with the same sine function value

We know that \(\sin\theta=\sin(180^{\circ}-\theta)\) for angles in the second quadrant. Given \(\theta = 190^{\circ}\), which is in the third quadrant. The reference angle for \(190^{\circ}\) is \(190^{\circ}- 180^{\circ}=10^{\circ}\).
The angle with the same sine value as \(190^{\circ}\) (since \(\sin\alpha=\sin(180^{\circ}+\beta)\) and \(\sin\alpha=\sin(360^{\circ}-\beta)\) in terms of sign and reference - angle, and we want \(0^{\circ}<\theta<360^{\circ}\)) is \(180^{\circ}+ 10^{\circ}=190^{\circ}\) (itself) and \(360^{\circ}-10^{\circ}=350^{\circ}\). But using the identity \(\sin\theta=\sin(180^{\circ}-\theta)\) in a different way (considering the unit - circle symmetry for sine values), the angle is \(180^{\circ}+10^{\circ}=190^{\circ}\) (not valid as we need a non - equal angle) and \(180^{\circ}-(- 10^{\circ})\) (using the fact that \(\sin\theta=\sin(180^{\circ}-\theta)\) where for \(\theta = 190^{\circ}\), the equivalent angle is \(180^{\circ}+10^{\circ}\) and \(360^{\circ}-10^{\circ}\). The valid non - equal angle is \(350^{\circ}\) (because \(\sin190^{\circ}=\sin(360^{\circ}-10^{\circ})\), and \(\sin x=\sin(360^{\circ}-x)\) for \(x = 10^{\circ}\)).

Step2: Find the angle with the same cosine function value

We know that \(\cos\theta=\cos(360^{\circ}-\theta)\). The reference angle for \(190^{\circ}\) is \(190^{\circ}-180^{\circ}=10^{\circ}\).
Using the identity \(\cos\alpha=\cos(360^{\circ}-\alpha)\), if \(\alpha = 190^{\circ}\), then \(\cos190^{\circ}=\cos(360^{\circ}-190^{\circ})\). So \(\theta = 170^{\circ}\) (because \(\cos190^{\circ}=\cos(180^{\circ}+10^{\circ})=-\cos10^{\circ}\) and \(\cos170^{\circ}=\cos(180^{\circ}-10^{\circ})=-\cos10^{\circ}\))

Answer:

For the sine function value: \(350\) degrees
For the cosine function value: \(170\) degrees