Use a double or half angle identity to find the exact value of the following expression: if cos x= 4/5 and 270° < x < 360°, find sin 2x
Please help! Thanks
cosx = 4/5 and x is in quad IV
So sinx = - 3/5
sin 2x = 2sinxcosx
= 2(-3/5)(4/5) = -24/25
if tan x = -7/24 and 3π/2 < x <2π find cot x/2
Thanks so much!
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if tanx = -7/24 and x is in IV
then sin x = -7/25 and cosx = 24/25
cot x/2
= cos x/2 / sin x/2
recall cos 2A = 2cos^2 A - 1 or 1 - 2sin^2 A
cos x = 2cos^ x/2 - 1
24/25 + 1 = 2cos^2 x/2 = 49/25
cos^2 x/2 = 49/50
cos x/2 = -7/(5√2) , if x is in IV , then x/2 is in II , therefore negative
cos x = 1 - 2sin^2 x/2
24/25 - 1 = -2sin^2 x/2 = -1/25
sin^2 x/2 = 1/50
sin x/2 = +1/(5√2)
cot x/2 = cos x/2 / sin x/2
= -7/(5√2) / 1/(5√2) = -7
Thank you soo much!
To find the value of sin 2x, we can use the double angle identity for sine:
sin 2x = 2sin x cos x
Given that cos x = 4/5, we need to find the value of sin x in the given range (270° < x < 360°).
First, let's find sin x using the Pythagorean identity:
sin^2 x + cos^2 x = 1
Since cos x = 4/5, we have:
sin^2 x + (4/5)^2 = 1
sin^2 x + 16/25 = 1
sin^2 x = 1 - 16/25
sin^2 x = 9/25
Taking the square root of both sides, we get:
sin x = ±√(9/25) = ±3/5
Since x is in the fourth quadrant (270° < x < 360°), sin x is negative:
sin x = -3/5
Now, let's substitute the values of sin x and cos x into the double angle identity:
sin 2x = 2sin x cos x
= 2(-3/5)(4/5)
= -24/25
Therefore, the exact value of sin 2x when cos x = 4/5 and 270° < x < 360° is -24/25.