Use the double angle identities to find sin2x if sinx= 1/square of 17 and cosx<0 . Give exact answer.
Note: if sin(x)>0 and cos(x)<0, the angle x is in the second quadrant.
therefore
cos(x)=-sqrt(1-sin²(x))
Calculate sin(2x) from the double angle formula:
sin(2x)=2sin(x)cos(x)
Since 2x lies in the third or fourth quadrant, sin(2x) is expected to be negative.
In a triagle abc angle b is 3 times angle a and angle c is 19 less than 6 times angle a.
What is the size of number a,b, and c
To find sin2x using the double angle identities, we will need to make use of the given value for sinx and the fact that cosx<0.
The double angle identity for sin2x is given as sin2x = 2sinxcosx.
We are given that sinx = 1/sqrt(17) and cosx < 0. Let's substitute these values into the double angle identity:
sin2x = 2 * sinx * cosx
Plugging in the given values:
sin2x = 2 * (1/sqrt(17)) * (cosx)
Since we are given that cosx < 0, we know that cosx is negative. Therefore, we can simplify the expression as follows:
sin2x = 2 * (1/sqrt(17)) * (-cosx)
Now, we need to determine the value of cosx. Given that sinx = 1/sqrt(17), we can use the Pythagorean identity to find cosx:
cosx = sqrt(1 - sin^2(x))
cosx = sqrt(1 - (1/sqrt(17))^2)
cosx = sqrt(1 - 1/17)
cosx = sqrt(16/17)
cosx = 4/sqrt(17)
Substituting this value back into the expression for sin2x, we get:
sin2x = 2 * (1/sqrt(17)) * (-4/sqrt(17))
Simplifying further:
sin2x = -8/17
Therefore, the exact value of sin2x, given sinx = 1/sqrt(17) and cosx < 0, is -8/17.
since sine is positive and cosine is negative , x must be in quadrant II
make a diagram in the second quadrant with opposite as 1 and hypotenuse as 17, so the adjacent is 4 by Pythagoras, and
cos x = -4/√17
sin 2x = 2sinxcosx = 2(1/√17)(-4/√17) = -8/17