Given that sin(x + 60) degree = cos(2x) degree, find tan(x + 60) degree.
Cos = sin(90 - 2x)
Sin(x + 60) = sin(90 - 2x)
x + 60 = 90 - 2 x
3x + 60 = 90
3x = 30
x = 10
Tan(x + 60)
tan(10 + 60)
tan(70)
= 2.747
well,
tan(x+60) = sin(x+60)/cos(x+60)
= cos(2x)/cos(x+60)
sin(x+60)-cos(2x) = 0
cos(30-x)-cos(2x) = 0
using the sum-to-product formulas,
2sin(x/2 + 15)sin(3x/2 + 15) = 0
set each factor to 0 to find values of x that make it zero.
Then take tan(x+60)
boy - how did I miss that solution?
of course, there are probably solutions in other quadrants, since sin(180-x) = sin(x)
not understood
Iam not understanding
To find tan(x + 60) degrees, we can use the identity tan^2(x) = 1 - cos^2(x).
First, let's rewrite the given equation using the cosine double-angle identity:
sin(x + 60) = cos(2x)
sin(x + 60) = cos^2(x) - sin^2(x)
Now, substitute sin^2(x) = 1 - cos^2(x) into the equation:
sin(x + 60) = cos^2(x) - (1 - cos^2(x))
sin(x + 60) = 2cos^2(x) - 1
Next, we can express cos^2(x) in terms of tan(x) using the identity cos^2(x) = 1 / (1 + tan^2(x)):
sin(x + 60) = 2(1 / (1 + tan^2(x))) - 1
sin(x + 60) = (2 - 2tan^2(x)) / (1 + tan^2(x))
Finally, we can find tan(x + 60) by taking the square root of both sides of the equation:
tan(x + 60) = sqrt((2 - 2tan^2(x)) / (1 + tan^2(x)))
This gives the expression for tan(x + 60) in terms of tan(x).