Givens: cos(q) = 3/5
Sin(2q) = 2 sin(q)*cos(q)
Step One
Find sin(q)
Sin(q)= sqrt(1 - cos^2(q) )
Sin(q) = sqrt(1 - (3/5)^2 ) = sqrt(1 - (9/25) = sqrt( (16/25) ) = 4/5. Because the angle is in the first quadrant, sin(q) > 0.
Step 2
Find the value of sin(2q)
sin(2q) = 2 sin(q) * cos(q)
Sin(2q) = 2* 3/5 * 4/5
sin(2q) = 24/25.
Note: you should notice that sin(2q) is less than 1 which is the way a sine function should behave.
Hello!
To find sin 2q, use the formula sin 2q = 2sin(q) cos(q). Also, since theta is in the first quadrant, all the trigonometric ratios are positive.
Since the question has only given the value of cos q = 3/5, we need to find sin q.
cos q = x / r and sin q = y / r. To find y, use the Pythagorean Theorem.
3² + y² = 5²
9 + y² = 25 (subtract 9 from both sides)
y² = 16 (square root both sides of the equation)
y = 4 | sin(q) = 4 / 5
Then, we can substitute these values into sin2(q) = 2sin(q) cos(q).
sin 2q = 2(4/5)(3/5)
sin 2q = (8/5)(3/5)
sin 2q = 24/25
The exact value of sin 2q is 24/25.
