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Please help me to solve this problem on how to proof if it is a double angle

$$\begin{align*} \cos 2 A &= 1-2\sin^2A\\ \sin 2 A &= 2\sin A\cos A\\ \tan 2A &=2\sin A\cos A\\ \end{align*}$$

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closed as off-topic by Integrator, Davide Giraudo, Souvik Dey, Najib Idrissi, tetori Dec 14 at 14:18

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do you know the formula of $\sin(A+B), \cos (A+B)$ ? –  Santosh Linkha Jan 10 '13 at 23:03
Do you have access to the angle sum identities? If so, start with an angle sum of an angle with itself. (e.g., $\cos(\alpha + \alpha)$. –  Todd Wilcox Jan 10 '13 at 23:04
@user335160: Welcome to MSE! Is this homework? If so, it should be tagged as such. What have you tried? Where are you confused? These sorts of additions to questions help to MSE community to better help you with questions. Regards! –  Amzoti Jan 10 '13 at 23:05
sorry I don't know because I am new to this. please help me to solve this problem with solution. thank you very much –  user335160 Jan 10 '13 at 23:06
It would help us help you if you answer the first two questions. –  amWhy Jan 10 '13 at 23:07

1 Answer 1

Exists two basic identities for sum of angles and they are $$\sin(A+B)=\sin A\cos B+\sin B\cos A$$ $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ if in these you put $A=B$ you get $$\sin(2A)=2\sin A\cos A$$ $$\cos(2A)=\cos^2 A-\sin^2 A$$ then you need to manipulate following identities $$\tan A=\frac{\sin A}{\cos A}$$ $${\sin^2 A}+{\cos^2 A}=1$$

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