Respuesta :

jacob193

Answer:

[tex]\displaystyle \frac{\cos x}{1 + \sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)[/tex].

Overview of the steps:

  • Apply the double-angle identity of sines and cosines to the left-hand side of the equation.
  • Apply the Pythagorean identity to the left-hand side of the equation.
  • Apply the angle sum and difference identity of sines and cosine to the right-hand side of the equation.

Step-by-step explanation:

Double-angle identity of sines and cosines:

  • [tex]\cos (2\,\alpha) = \cos^2\alpha - \sin^2\alpha = (\cos\alpha + \sin\alpha)\, (\cos\alpha - \sin\alpha)[/tex].
  • [tex]\sin(2\,\alpha) = 2\, \sin\alpha\, \cos\alpha[/tex].

Pythagorean identity for the sine and cosine of the same angle:

[tex]1 = \cos^2\alpha + \sin^2\alpha[/tex].

Angle sum and difference identity of sines and cosines:

[tex]\sin(\alpha - \beta) = \sin\alpha\, \cos\beta - \cos\alpha \, \sin\beta[/tex].

[tex]\cos(\alpha - \beta) = \cos\alpha\, \cos\beta + \sin\alpha \, \sin\beta[/tex].

Consider [tex]x[/tex] as the sum of two angles of size [tex](x/2)[/tex]. Start by applying the double-angle identity to the left-hand side.

[tex]\begin{aligned} \text{L.H.S.}&= \frac{\cos(x)}{1 + \sin(x)} \\ &= \frac{\cos^2(x / 2) - \sin^2(x / 2)}{1 + 2\, \sin(x / 2)\, \cos(x / 2)}\end{aligned}[/tex].

Apply the Pythagorean identity to rewrite the "1" in the denominator as [tex]\left(\cos^2(x / 2) + \sin^2(x / 2)\right)[/tex].

[tex]\begin{aligned} \text{L.H.S.}&= \frac{\cos(x)}{1 + \sin(x)} \\ &= \frac{\cos^2(x / 2) - \sin^2(x / 2)}{1 + 2\, \sin(x / 2)\, \cos(x / 2)}\\ &= \frac{\cos^2(x / 2) - \sin^2(x / 2)}{\sin^2(x/2) + 2\,\sin(x/2)\, \cos(x/2) + \cos^2 (x/2)}\end{aligned}[/tex].

Note that the denominator is now a perfect square. On the other hand, the numerator is in the form [tex](x^2 - y^2)[/tex], which is equal to [tex](x + y)\, (x - y)[/tex]. Rewrite and simplify this expression:

[tex]\begin{aligned} \text{L.H.S.}&= \frac{\cos(x)}{1 + \sin(x)} \\ &= \frac{\cos^2(x / 2) - \sin^2(x / 2)}{1 + 2\, \sin(x / 2)\, \cos(x / 2)}\\ &= \frac{\cos^2(x / 2) - \sin^2(x / 2)}{\sin^2(x/2) + 2\,\sin(x/2)\, \cos(x/2) + \cos^2 (x/2)} \\[1em] &= \frac{(\cos(x/2) + \sin(x/2))\, (\cos(x/2) - \sin(x/2))}{\left(\sin(x/2) + \cos(x/2)\right)^2} \\ &= \frac{\cos(x/2) - \sin(x/2)}{\sin(x/2) + \cos(x/2)}\end{aligned}[/tex].

The tangent of an angle is equal to the ratio between its sine and its cosine. Apply the angle sum and difference identity of sine and cosine to the right-hand side.

Note, that the sine and cosine of [tex](\pi/4)[/tex] are both equal to [tex]\left(\sqrt{2}/2\right)[/tex].

[tex]\begin{aligned}\text{R.H.S.} &= \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \\ &= \frac{\sin((\pi/4) - (x/2))}{\cos((\pi/4) - (x/2))}\\ &= \frac{\left(\sqrt{2}/2\right)\, \cos(x/2) - \left(\sqrt{2}/2\right)\, \sin(x/2)}{\left(\sqrt{2}/2\right)\, \cos(x/2) + \left(\sqrt{2}/2\right)\, \sin(x/2)} \\ &= \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}\end{aligned}[/tex].

Therefore:

[tex]\displaystyle \text{L.H.S.} = \frac{\cos(x/2) - \sin(x/2)}{\sin(x/2) + \cos(x/2)} =\text{R.H.S.}[/tex].

[tex]\displaystyle \frac{\cos x}{1 + \sin x} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)[/tex].