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The angle 01 is located in Quadrant IV and cos (01) = 3/5

The angle 01 is located in Quadrant IV and cos (01) = 3/5-example-1

2 Answers

4 votes

Answer:

4/5

Explanation:

Using the Pythagorean Identity,


\sin {}^(2) ( \alpha ) + \cos {}^(2) ( \alpha ) = 1

Note that


\cos {}^(2) ( \alpha )

is basically


( \cos( \alpha ) ) {}^(2)

So that means


\cos {}^(2) ( \alpha ) = ( (3)/(5) ) {}^(2) = (9)/(25)

So we have


\sin {}^(2) ( \alpha ) + (9)/(25) = 1

Solve for sin a.


\sin {}^(2) ( \alpha ) = (25)/(25) - (9)/(25)

25/25 is the same as 1.


\sin {}^(2) ( \alpha ) = (16)/(25)

Take the square root.

since sin is positive in the first quadrant, take the principal square root.


\sin( \alpha ) = ( √(16) )/( √(25) )


\sin( \alpha ) = (4)/(5)

User Mancze
by
5.0k points
3 votes

The value of sin(θ1) is -4/5.

To solve for the value of sin(θ1) when angle θ1 is located in Quadrant IV and cos(θ1) = 3/5:

Use the Pythagorean identity: We know that sin²(θ) + cos²(θ) = 1 for any angle θ. So, we can rearrange this equation to solve for sin(θ): sin²(θ) = 1 - cos²(θ). In this case, we are given that cos(θ1) = 3/5, so we can plug that value into the equation to get: sin²(θ1) = 1 - (3/5)² = 16/25.

Take the square root of both sides: Since we want to solve for sin(θ1), and sin(θ) can be positive or negative depending on the quadrant, we need to take the square root of both sides.

The sine function is negative in Quadrant IV, so we need to take the negative square root to get the correct value. Therefore, sin(θ1) = -√(16/25) = -4/5.

Therefore, the value of sin(θ1) is -4/5.

User Carrabino
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5.3k points