Question

If cos of theta = 3/5 and q terminates in the first quadrant, find the exact value of sin 2q.

q=theta btw

Answer 1

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.

Answer 2

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.