2
$\begingroup$

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 $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$

$\endgroup$
5
$\begingroup$

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*}

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

OR

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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!

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.