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Given that cos(θ)=3√10, and θ is in quadrant iv, what is sin(2θ)?

2 Answers

7 votes

Final answer:

The value of sin(2θ) is -6√890.

Step-by-step explanation:

Given that cos(θ) = 3√10 and θ is in quadrant IV, we can use the trigonometric identity sin(2θ) = 2sin(θ)cos(θ). Since cos(θ) is given and positive in quadrant IV, sin(θ) will be negative. We can find sin(θ) using the Pythagorean identity sin(θ) = √(1 - cos^2(θ)).



First, we can find sin(θ):



sin(θ) = √(1 - (3√10)^2)



sin(θ) = √(1 - 90)



sin(θ) = √(-89)



Since θ is in quadrant IV, sin(θ) is negative:



sin(θ) = -√89



Now, we can find sin(2θ):



sin(2θ) = 2sin(θ)cos(θ)



sin(2θ) = 2(-√89)(3√10)



sin(2θ) = -6√890

User Alessandro Benoit
by
8.1k points
1 vote

Final answer:

Using the double-angle formula for sine and the given cosine value, we calculate sin(2θ) to be -2√{21}/10, considering that θ is in the fourth quadrant where sine is negative.

Step-by-step explanation:

To find sin(2θ) given that cos(θ) = √{­3/10} and θ is in the fourth quadrant, we can use the double-angle formula for sine, which is sin(2θ) = 2sin(θ)cos(θ). However, first we need to find sin(θ). Since θ is in the fourth quadrant, sin(θ) must be negative and we can use the Pythagorean identity sin²(θ) + cos²(θ) = 1 to find sin(θ).

cos²(θ) = (√{3/10})² = 3/10

sin²(θ) = 1 - cos²(θ) = 1 - 3/10 = 7/10

sin(θ) = -√{7/10}

Now, apply the double-angle formula:

sin(2θ) = 2sin(θ)cos(θ) = 2(-√{7/10})(√{3/10})

sin(2θ) = -2√{21}/10

User Vishnu Vinod
by
8.2k points
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