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PROVE THAT

Posted October 28th, 2009 by alpha
in

Given :$ sin\alpha+sin\beta=a $ and $ cos\alpha+cos\beta=b $
then prove that $ sin(\alpha+\beta)=\frac{2ab}{a^2+b^2 } $

‹ Solve the following equation: Trigonometric equation ›

Comments

solution

On October 28th, 2009 Structure says:

We have:
$ a = sin \alpha + sin \beta = 2 sin(\frac{\alpha+\beta}{2}) cos(\frac{\alpha−\beta}{2}) $
$ b = cos \alpha + cos \beta = 2 cos(\frac{\alpha+\beta}{2}) cos(\frac{\alpha−\beta}{2}) $

So:

$$\frac{a}{b} = \frac{2 sin(\frac{\alpha+\beta}{2}) cos(\frac{\alpha−\beta}{2})}{2 cos(\frac{\alpha+\beta}{2}) cos(\frac{\alpha−\beta}{2})}= \tan(\frac{\alpha+\beta}{2})$$

If we replace $ \frac{a}{b} $ we get:

$$\frac{2ab}{a^2 + b^2} = \frac{2\frac{a}{b}}{1 + (\frac{a}{b})^2}= \frac{2 \tan(\frac{\alpha+\beta}{2})}{1 + ( \tan(\frac{\alpha+\beta}{2}) )^2}$$

but we know that:

$$\sin{t}=\frac{2\tan{\frac{t}{2}}}{1+(\tan{\frac{t}{2}})^2}$$

so:

$$\sin(\alpha+\beta)=\frac{2 \tan(\frac{\alpha+\beta}{2})}{1 + ( \tan(\frac{\alpha+\beta}{2}) )^2}$$

in the end:

$$\sin(\alpha+\beta)=\frac{2ab}{a^2 + b^2}$$

maths

On December 2nd, 2009 csmuthu says:

sin^2a+sin^2b+sin^2c=1-2sina.sinb.sinc

~~THANK YOU ~~

On November 2nd, 2009 alpha says:

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

''ALPHA" =)

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