a)cos^2 alfa - sin^2 alfa = 1 +2sin^2 alfa

b)cos^2 alfa - sin^2 alfa =2cos^2 alfa - 1

c)1+tan^2 alfa = 1/sin^2 alfa

e)1+tan^2 alfa /tan^2 alfa = 1/sin^2 alfa

help!(Headbang)

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- Mar 26th 2010, 09:33 AMlebanonestablish the following:
a)cos^2 alfa - sin^2 alfa = 1 +2sin^2 alfa

b)cos^2 alfa - sin^2 alfa =2cos^2 alfa - 1

c)1+tan^2 alfa = 1/sin^2 alfa

e)1+tan^2 alfa /tan^2 alfa = 1/sin^2 alfa

help!(Headbang) - Mar 26th 2010, 11:12 AMGrandad
Hello lebanonYou have one or two mistakes in these equations, which I'll point out as we come to them. They are all based on the fundamental relationship between sine and cosine:

$\displaystyle \cos^2\alpha+\sin^2\alpha = 1$ ...(1)which is the trigonometric form of Pythagoras' Theorem.

From this we can deduce:$\displaystyle \cos^2\alpha=1-\sin^2\alpha $ ...(2)and:$\displaystyle \sin^2\alpha=1-\cos^2\alpha $ ...(3)So in (a), we start with the LHS and use (2) to eliminate $\displaystyle \cos^2\alpha$:

$\displaystyle \cos^2\alpha - \sin^2\alpha = (1-\sin^2\alpha) - \sin^2\alpha$In (b) we do the same, but eliminating $\displaystyle \sin^2\alpha$ using (3):$\displaystyle = 1- 2\sin^2\alpha$ (Note the minus sign.)

$\displaystyle \cos^2\alpha - \sin^2\alpha = \cos^2\alpha -(1-\cos^2\alpha)$In (c) there's another mistake: the RHS should be $\displaystyle \frac{1}{\cos^2\alpha}$. This time, we start with (1) and divide both sides by $\displaystyle \cos^2\alpha$:$\displaystyle = 2\cos^2\alpha-1$

$\displaystyle \cos^2\alpha+\sin^2\alpha = 1$Finally, for part (e) (I wonder what happened to (d)?) we take the result in part (c) and divide both sides by $\displaystyle \tan^2\alpha$:

$\displaystyle \Rightarrow 1+ \frac{\sin^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\ alpha}$

$\displaystyle \Rightarrow 1+ \tan^2\alpha=\frac{1}{\cos^2\alpha}$, because $\displaystyle \tan\alpha = \frac{\sin\alpha}{\cos\alpha}$$\displaystyle 1+ \tan^2\alpha=\frac{1}{\cos^2\alpha}$Grandad

$\displaystyle \Rightarrow \frac{1+ \tan^2\alpha}{\tan^2\alpha}=\frac{1}{\cos^2\alpha\ tan^2\alpha}$$\displaystyle =\frac{1}{\sin^2\alpha}$, again using the fact that $\displaystyle \tan\alpha = \frac{\sin\alpha}{\cos\alpha}$, so $\displaystyle \cos\alpha\tan\alpha = \sin\alpha$.