It isn't clear what you want to solve for. To solve trigonometric equations, it often helps to convert other angular functions (tangent, cotangent, secant, cosecant) into the equivalent of sines and cosines. However, the details of course depend on the specific case.
Trig identity... sin/cos = tangent
In a right triangle, its Opposite/Hypotenuse I always use: Soh (sin, opposite/hypotenuse) Cah (cosine, adjacent/hypotenuse) Toa (tangent, opposite/adjacent) Hope this helped! :)
The inverse of the cosine is the secant.
If X and Y are sides of a right triangle, R is the hypoteneuse, and theta is the angle at the X-R vertex, then sin(theta) is Y / R and cosine(theta) is X / R. It follows, then, that X is R cosine(theta) and Y is R sin(theta)
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
Tangent = sine/cosine provided that cosine is non-zero. When cosine is 0, then tangent is undefined.
you can use the sine, cosine, tangent formula.
Sine = -0.5 Cosine = -0.866 Tangent = 0.577
Reciprocal of tangent is '1 /tangent' or ' Cosine / Sine '
Yes, sine, cosine, tangent definitions are based on right triangles
It means for any right angle triangle:- sine = opposte/hypotenuse cosine = adjacent/hypotenuse tangent = opposite/adjacent
Cotangent is 1 / tangent. Since tangent is sine / cosine, cotangent is cosine / sine.
Cotangent is ' 1/tangent' or ' Cosine / Sine'.
in trigonometry
Trigonometry
No, it does not.
The cosine is ±1/sqrt(5) = ±0.4472 (approx).