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Prove that

Posted October 28th, 2009 by alpha
in

GIVEN:

$$2tan\frac{\alpha}{2}=tan\frac{\beta}{2}$$

Then prove that:

$$cos\alpha=\frac{3+5cos\beta}{5+3cos\beta}$$
‹ Trigonometric equation Find the value of : ›

Comments

solution

On October 28th, 2009 Structure says:
$$x = \tan\frac{\alpha}{2}$$
$$y = \tan\frac{\beta}{2}$$
$$y = 2x$$
$$cos\alpha = \frac{1 - x^2}{1 + x^2}$$
$$cos\beta = \frac{1 - y^2}{1 + y^2} = \frac{1 - 4x^2}{1 + 4x^2}$$
$$\frac{3+5cos\beta}{5+3cos\beta} = \frac{3+5\frac{1 - 4x^2}{1 + 4x^2}}{5+3\frac{1 - 4x^2}{1 + 4x^2}} = \frac{3 + 12 x^2+5 - 20x^2}{5+20x^2+3 - 12x^2} = \frac{8 - 8x^2}{8 + 8^2} = \frac{1 - x^2}{1 + x^2} = \cos \alpha$$

~~THANK YOU ~~

On November 2nd, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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