By definition, tan x = opposite/adjacent in a right triangle.
Since sin x = opposite/hypotenuse and cos x = adjacent/hypotenuse,
tan x = (opposite/hypotenuse) / (adjacent/hypotenuse) = opposite/adjacent = sin x / cos x
cot x is the reciprocal of tan x.
Since tan x = sin x / cos x,
cot x = 1 / tan x = 1 / (sin x / cos x) = cos x / sin x
By definition, sec x is the reciprocal of cos x.
In a right triangle, cos x = adjacent/hypotenuse,
so sec x = hypotenuse/adjacent = 1 / cos x
By definition, csc x is the reciprocal of sin x.
In a right triangle, sin x = opposite/hypotenuse,
so csc x = hypotenuse/opposite = 1 / sin x
From the Pythagorean theorem in a unit circle:
For any point (cos x, sin x) on the unit circle,
(cos x)² + (sin x)² = 1 (since radius = 1)
Thus, sin²x + cos²x = 1
Start with sin²x + cos²x = 1
Divide both sides by sin²x:
(sin²x/sin²x) + (cos²x/sin²x) = 1/sin²x
1 + cot²x = csc²x
Rearrange: csc²x - cot²x = 1
Start with sin²x + cos²x = 1
Divide both sides by cos²x:
(sin²x/cos²x) + (cos²x/cos²x) = 1/cos²x
tan²x + 1 = sec²x
Rearrange: sec²x - tan²x = 1
Start with cos(2x) = 2cos²x - 1
Solve for cos²x:
2cos²x = 1 + cos(2x)
cos²x = (1 + cos(2x)) / 2
Start with cos(2x) = 1 - 2sin²x
Solve for sin²x:
2sin²x = 1 - cos(2x)
sin²x = (1 - cos(2x)) / 2
Using sin(A+B) = sinA cosB + cosA sinB
Set A = x and B = x:
sin(x+x) = sinx cosx + cosx sinx
sin(2x) = 2 sinx cosx
Using cos(A+B) = cosA cosB - sinA sinB
Set A = x and B = x:
cos(x+x) = cosx cosx - sinx sinx
cos(2x) = cos²x - sin²x
Start with cos(2x) = cos²x - sin²x
Replace cos²x with (1 - sin²x) from sin²x + cos²x = 1:
cos(2x) = (1 - sin²x) - sin²x
cos(2x) = 1 - 2sin²x
Start with cos(2x) = cos²x - sin²x
Replace sin²x with (1 - cos²x) from sin²x + cos²x = 1:
cos(2x) = cos²x - (1 - cos²x)
cos(2x) = cos²x - 1 + cos²x
cos(2x) = 2cos²x - 1
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