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While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$. Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$

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The identity is true. \begin{align*} \cos x + \sin x \tan \frac{x}{2} & =(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})+(2\sin \frac{x}{2}\cos \frac{x}{2})(\frac{\sin \frac{x}{2}}{\cos\frac{x}{2}})\\ & = \cos^2\frac{x}{2}+\sin^2 \frac{x}{2}\\ & = 1 \end{align*}

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    $\begingroup$ To produce $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, $\cot x$, type \sin x, \cos x, \tan x, \csc x, \sec x, \cot x, respectively, in math mode. $\endgroup$ – N. F. Taussig Aug 24 '15 at 11:24
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$$\sin2y\cdot\tan y=2\sin y\cos y\cdot\dfrac{\sin y}{\cos y}=1-\cos2y$$

OR

$$\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=\tan y$$

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Yup! We have $$\cos x + \sin x \tan \frac{x}{2} = (\cos ^2 \frac{x}{2} - \sin ^2 \frac{x}{2}) + (2 \sin \frac{x}{2} \cos \frac{x}{2}) \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \cos ^2 \frac{x}{2} + \sin ^2 \frac{x}{2} = 1$$

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if $\tan\frac x2=t,$ then $\cos x =\frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$ so it works!

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Let me try. You have: $$\cos x + \sin x \tan (\frac{x}{2}) = 1 - 2\sin^2 (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) \tan (\frac{x}{2}) = 1 -2\sin^2 (\frac{x}{2}) + 2\sin^2 (\frac{x}{2}) = 1.$$

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Notice, $$LHS=\cos x+\sin x\tan \frac{x}{2}$$ $$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\tan \frac{x}{2}$$ $$=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{2\tan^2 \frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ $$=\frac{1-\tan^2\frac{x}{2}+2\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ $$=\frac{1+\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}=1=RHS$$

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We have $$\cos\left(x\right)+\sin\left(x\right)\tan\left(\frac{x}{2}\right)=1\Leftrightarrow\cos\left(x\right)\cos\left(\frac{x}{2}\right)+\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\cos\left(\frac{x}{2}\right) $$ and, using product to sum identity and the fact that cos is a even function$$\cos\left(x\right)\cos\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)+\frac{1}{2}\cos\left(\frac{3x}{2}\right) $$ and again using product to sum identity we get $$\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\frac{1}{2}\cos\left(\frac{x}{2}\right)-\frac{1}{2}\cos\left(\frac{3x}{2}\right) $$ and we have done.

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An easy way to do that is: sinx = sqrt((1+cosx)(1-cosx)) so LHS becomes: cosx + sqrt((1+cosx)(1-sinx))sqrt((1-cosx)/1+cosx)) cosx + 1 - cosx = 1 = RHS

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$ t = \tan(../2) $

$$ \sin ( ..) = \frac{ 2 t }{1+t^2} ; \, \cos ( ..) = \frac{ 1-t^2 } {1+t^2} $$

Plug in $$\cos x+\sin x\tan \frac{x}{2} \rightarrow 1 $$

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