2

Know that $\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}$ calculate $\sin\alpha$

My proof:
$\tan\left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\\ \frac{\sin\left(\alpha-\frac{\pi}{4}\right)}{\cos\left(\alpha-\frac{\pi}{4}\right)}=\frac{1}{3}\\3\sin\left(\alpha-\frac{\pi}{4}\right)=\cos\left(\alpha-\frac{\pi}{4}\right)\\\sin^2\left(\alpha-\frac{\pi}{4}\right)+9\sin^2\left(\alpha-\frac{\pi}{4}\right)=1\\\sin\left(\alpha-\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt{10}}\\ \sin\left(\alpha-\frac{\pi}{4}\right)=\sin\alpha\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\alpha=\frac{\sqrt{2}}{2}\sin\alpha-\frac{\sqrt2}{2}\cos\alpha=\frac{\sqrt2}{2}\left(\sin\alpha-\cos\alpha\right)=\pm\frac{1}{\sqrt{10}}\\\sin\alpha-\cos\alpha=\pm\frac{1}{\sqrt{5}}\\\sin\alpha=\pm\frac{1}{\sqrt{5}}+\cos\alpha$
I have no idea how to determine $\sin\alpha$

vmahth1
  • 479

3 Answers3

6

$\tan (\alpha -\frac {\pi} 4)=\frac {\tan (\alpha)-1} {1+\tan \alpha}=\frac 1 3$ and this gives $\tan \alpha =2$. Can you find $\sin \alpha$ from this?

Kavi Rama Murthy
  • 311,013
  • Yes, thank you. $\tan\alpha=2\Rightarrow \cos\alpha=\frac{1}{2}\sin\alpha\ \sin^2\alpha+\frac{1}{4}\sin^2\alpha=1\Rightarrow\sin\alpha=\pm\frac{2}{\sqrt{5}}$ – vmahth1 Nov 24 '19 at 05:39
2

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$ and $-\arctan(x)=\arctan(-x),$

$$\alpha=n\pi+\dfrac\pi4-\arctan\dfrac13=n\pi+\arctan\dfrac{1+\left(-\dfrac13\right)}{1-1\left(-\dfrac13\right)}=n\pi+\arctan\dfrac12$$ where $n$ is any integer

$$\sin\alpha=(-1)^n\sin\left(\arctan\dfrac12\right)$$

If $\arctan\dfrac12=y,\tan y=\dfrac12,$

$$\dfrac{\sin y}1=\dfrac{\cos y}2=\pm\sqrt{\dfrac{\sin^2y+\cos^2y}{2^2+1^2}}$$

Now as $0<\arctan\dfrac12=y<\dfrac\pi2,\sin y>0,\cos y>0$

lab bhattacharjee
  • 274,582
0

We have $\tan\left(\dfrac\pi4-\alpha\right)=\dfrac{1-\tan\alpha}{1+\tan\alpha}=\dfrac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$

$$\dfrac{\sin\alpha-\cos\alpha}1=\dfrac{\sin\alpha+\cos\alpha}3=\pm\sqrt{\dfrac{(\sin\alpha-\cos\alpha)^2+(\sin\alpha+\cos\alpha)^2}{1^2+3^2}}$$

$$\dfrac{\sin\alpha-\cos\alpha}1=\dfrac{\sin\alpha+\cos\alpha}3=\pm\sqrt{\dfrac2{1^2+3^2}}=\dfrac{\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha}{1+3}$$

lab bhattacharjee
  • 274,582