If 13sinx+5=0 and x�[0;270],determine witout the use of a calculator,the value of sin2x
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13sinx+5=0
sinx = -5/13
sinx is negative in QIII,QIV, but we want the QIII value.
so, cosx = -12/13
sin2x = 2sinx cosx = 2(-5/13)(-12/13) = 120/169
To determine the value of sin(2x) without a calculator, we need to first find the value of x that satisfies the equation 13sin(x) + 5 = 0 within the given interval [0, 270].
Let's solve the equation:
13sin(x) + 5 = 0
Subtracting 5 from both sides:
13sin(x) = -5
Next, divide both sides by 13:
sin(x) = -5/13
Now, to find the value of sin(2x), we can use the double angle identity for sine:
sin(2x) = 2sin(x)cos(x)
The value of sin(x) is given as -5/13, so we need to find cos(x) to calculate sin(2x).
To find cos(x), we can use the Pythagorean identity for trigonometric functions:
sin^2(x) + cos^2(x) = 1
Substituting sin(x) = -5/13:
(-5/13)^2 + cos^2(x) = 1
Simplifying:
25/169 + cos^2(x) = 1
cos^2(x) = 1 - 25/169
cos^2(x) = 144/169
cos(x) = sqrt(144/169)
cos(x) = 12/13
Now, we have both sin(x) and cos(x).
Substituting these values into sin(2x) = 2sin(x)cos(x):
sin(2x) = 2 * (-5/13) * (12/13)
Multiplying:
sin(2x) = -120/169
Therefore, the value of sin(2x) is -120/169.