MHB Proving Trig Identity: (Sin2x-tanx)/cos2x=tanx

  • Thread starter Thread starter paix1988
  • Start date Start date
  • Tags Tags
    Identity Trig
Click For Summary
The discussion focuses on proving the trigonometric identity (Sin2x - tanx) / cos2x = tanx. Participants suggest using the identities sin 2x = 2 sin x cos x and tan x = sin x / cos x to simplify the right-hand side. The next step involves combining the terms in the numerator into a single fraction with a common denominator. Factorization of the resulting expression is recommended to facilitate further simplification. The goal is to demonstrate the equality through these algebraic manipulations.
paix1988
Messages
2
Reaction score
0
May someone kindly assist me to prove this trig identity

(Sin2x-tanx) / cos2x = tanx

Thank you for your assistance
 
Mathematics news on Phys.org
paix1988 said:
May someone kindly assist me to prove this trig identity

(Sin2x-tanx) / cos2x = tanx

Thank you for your assistance

Making use of the identities $\sin 2x = 2 \sin x \cos x$ and $\tan x = \sin x / \cos x$ we can write the RHS as:
$$\frac{2 \sin x \cos x -\frac{\sin x}{\cos x}}{\cos 2x}.$$
To proceed, write the numerator as one fraction (thus with denominator $\cos x$) and factorize!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...