Find the EXACT value of sin(2A) if cscA=5/4
Assume A is a Quadrant I angle in standard position.
sin(2A+=
draw a triangle that fits the condition.
cscA = 5/4 means
sinA = 4/5
cosA = 3/5
Now, recall your double-angle formula
sin(2A) = 2 sinA cosA
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To find the exact value of sin(2A), we can use a trigonometric identity.
The trigonometric identity for sin(2A) is: sin(2A) = 2sin(A)cos(A)
Given that csc(A) = 5/4, we can find the values of sin(A) and cos(A) using this information.
Since csc(A) is the reciprocal of sin(A), we have: sin(A) = 1/csc(A) = 4/5
Now, to find cos(A), we can use the Pythagorean identity: sin^2(A) + cos^2(A) = 1
Plugging in sin(A) = 4/5, we can solve for cos(A):
(4/5)^2 + cos^2(A) = 1
16/25 + cos^2(A) = 1
cos^2(A) = 1 - 16/25
cos^2(A) = 9/25
cos(A) = ± √(9/25) = ± 3/5
Now, we have the values of sin(A) = 4/5 and cos(A) = ± 3/5.
Plugging these values into the trigonometric identity for sin(2A), we get:
sin(2A) = 2sin(A)cos(A) = 2(4/5)(± 3/5)
Multiplying these values together, we get:
sin(2A) = 8/25 (± 3/5)
Therefore, the exact value of sin(2A) is: sin(2A) = 8/25 (± 3/5)