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The angle 
theta_(1) is located in Quadrant IV, and 
cos(theta_(1))=(3)/(5).
What is the value of 
sin(theta_(1)) ? Express your answer exactly.

sin(theta_(1))=

The angle θ1 \theta_{1} is located in Quadrant IV, and cos(θ1)=35 \cos \left(\theta_{1}\right)=\frac{3}{5} .\newlineWhat is the value of sin(θ1) \sin \left(\theta_{1}\right) ? Express your answer exactly.\newlinesin(θ1)= \sin \left(\theta_{1}\right)=

Full solution

Q. The angle θ1 \theta_{1} is located in Quadrant IV, and cos(θ1)=35 \cos \left(\theta_{1}\right)=\frac{3}{5} .\newlineWhat is the value of sin(θ1) \sin \left(\theta_{1}\right) ? Express your answer exactly.\newlinesin(θ1)= \sin \left(\theta_{1}\right)=
  1. Given Information: We are given that cos(θ1)=35\cos(\theta_{1}) = \frac{3}{5} and that θ1\theta_{1} is in Quadrant IV. In Quadrant IV, cosine is positive and sine is negative. We can use the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^{2}(\theta) + \cos^{2}(\theta) = 1 to find the value of sin(θ1)\sin(\theta_{1}).
  2. Solving for sin2(θ)\sin^2(\theta): First, we will solve for sin2(θ1)\sin^2(\theta_{1}). We know that cos2(θ1)=(35)2\cos^2(\theta_{1}) = \left(\frac{3}{5}\right)^2. cos2(θ1)=925\cos^2(\theta_{1}) = \frac{9}{25}
  3. Using Pythagorean Identity: Now, we use the Pythagorean identity to find sin2(θ1)\sin^2(\theta_{1}).
    sin2(θ1)=1cos2(θ1)\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})
    sin2(θ1)=1925\sin^2(\theta_{1}) = 1 - \frac{9}{25}
    sin2(θ1)=2525925\sin^2(\theta_{1}) = \frac{25}{25} - \frac{9}{25}
    sin2(θ1)=1625\sin^2(\theta_{1}) = \frac{16}{25}
  4. Finding sin(θ)\sin(\theta): Next, we take the square root of both sides to find sin(θ1)\sin(\theta_{1}). Since θ1\theta_{1} is in Quadrant IV, we must take the negative square root because sine is negative in this quadrant.\newlinesin(θ1)=1625\sin(\theta_{1}) = -\sqrt{\frac{16}{25}}\newlinesin(θ1)=45\sin(\theta_{1}) = -\frac{4}{5}

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