Sin(20-30°)=√3÷2for0°<0<180°
To prove that sin(20-30°) is equal to √3/2 for 0°<θ<180°, we can use the trigonometric identities and angles in the given range.
Step 1: Convert the angles to degrees:
20-30° = -10°
Step 2: Apply the angle addition formula for sine:
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
In this case, A = 20° and B = 30°, so we have:
sin(20-30°) = sin(20°)cos(30°) - cos(20°)sin(30°)
Step 3: Find the values of sin(20°), cos(30°), cos(20°), and sin(30°):
Using the unit circle or a calculator, we find:
sin(20°) ≈ 0.3420
cos(30°) = √3/2
cos(20°) ≈ 0.9397
sin(30°) = 1/2
Step 4: Substitute the values into the formula:
sin(20-30°) ≈ 0.3420 * (√3/2) - 0.9397 * (1/2)
Step 5: Simplify the expression:
sin(20-30°) ≈ 0.1710√3 - 0.46985
Step 6: Convert √3 to its decimal approximation:
√3 ≈ 1.7321
Step 7: Substitute the decimal approximation and simplify further:
sin(20-30°) ≈ 0.1710 * 1.7321 - 0.46985
sin(20-30°) ≈ 0.2961 - 0.46985
sin(20-30°) ≈ -0.17375
Now we can conclude that sin(20-30°) is approximately equal to -0.17375, which does not match the given expression √3/2. It seems there might be a mistake or typo in the question.