Question

sin theta is equal to 2 then thita = ?​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sin(\theta)=2$

$\bf{\mapsto\,\,If\,\,\theta\,\,is\,\,real\,:}$

Then, no value of θ is satisfied by the above equation, because or real θ, sin(θ) ∈ [-1, 1]

$\bf{\mapsto\,\,If\,\,\theta\,\,is\,\,complex\,:}$

$\sf{sin(\theta)=2}$

$\sf{\implies\,\dfrac{e^{\displaystyle\,i\,\theta}-e^{\displaystyle\,-i\,\theta}}{2\,i}=2}$

$\sf{\implies\,e^{\displaystyle\,i\theta}-e^{\displaystyle\,-i\theta}=4\,i}$

$\sf{\implies\,e^{\displaystyle\,i\theta}-\dfrac{1}{e^{\displaystyle\,i\theta}}=4\,i}$

$\sf{\implies\,\dfrac{e^{\displaystyle\,2i\theta}-1}{e^{\displaystyle\,i\theta}}=4\,i}$

$\sf{\implies\,e^{\displaystyle\,2i\theta}-1=4\,i\,e^{\displaystyle\,i\theta}}$

$\sf{\implies\,e^{\displaystyle\,2i\theta}-4\,i\,e^{\displaystyle\,i\theta}-1=0}$

$\sf{\implies\,\left(e^{\displaystyle\,i\theta}\right)^2-4\,i\,\left(e^{\displaystyle\,i\theta}\right)-1=0}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=\dfrac{-(-4i)\pm\sqrt{(-4i)^2-4\cdot1\cdot(-1)}}{2\cdot1}}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=\dfrac{4i\pm\sqrt{-16+4}}{2}}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=\dfrac{4i\pm\sqrt{-12}}{2}}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=\dfrac{4i\pm2\sqrt{3}\,i}{2}}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=2i\pm\,i\sqrt{3}}$

$\sf{\implies\,e^{\displaystyle\,i\theta}=\left(2+\sqrt{3}\right)i\,\,\,\,\,\,or\,\,\,\,\,\,e^{\displaystyle\,i\theta}=\left(2-\sqrt{3}\right)i}$

Taking logarithm

$\sf{\implies\,\ln\left\{e^{\displaystyle\,i\theta}\right\}=\ln\left\{\left(2+\sqrt{3}\right)i\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\ln\left\{e^{\displaystyle\,i\theta}\right\}=\ln\left\{\left(2-\sqrt{3}\right)i\right\}}$

$\sf{\implies\,i\theta=\ln\left\{\left(2+\sqrt{3}\right)i\right\}\,\,\,\,\,\,or\,\,\,\,\,\,i\theta=\ln\left\{\left(2-\sqrt{3}\right)i\right\}}$

Now, consider the proposition

$\bf{The\,\,\,logarithm\,\,\,of\,\,\,a\,\,\,complex\,\,\,number\,:}$

$\tt{\ln(z)=\ln\left(|z|\,e^{\displaystyle\,i(\alpha+2n\pi)}\right)}$

Where α is the argument of z and n ∈ $\mathbb{Z}$

$\tt{\implies\,\ln(z)=\ln|z|+\ln\left(e^{\displaystyle\,i(\alpha+2n\pi)}\right)}$

$\tt{\implies\,\ln(z)=\ln|z|+i(\alpha+2n\pi)}$

Now,

$\sf{\implies\,\theta=\dfrac{1}{i}\cdot\ln\left\{\left(2+\sqrt{3}\right)i\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\theta=\dfrac{1}{i}\cdot\ln\left\{\left(2-\sqrt{3}\right)i\right\}}$

$\sf{\implies\,\theta=\dfrac{i}{i^2}\cdot\ln\left\{\left(2+\sqrt{3}\right)i\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\theta=\dfrac{i}{i^2}\cdot\ln\left\{\left(2-\sqrt{3}\right)i\right\}}$

$\sf{\implies\,\theta=-i\cdot\ln\left\{\left(2+\sqrt{3}\right)i\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\theta=-i\cdot\ln\left\{\left(2-\sqrt{3}\right)i\right\}}$

$\sf{\implies\,\theta=-i\cdot\left[\ln\left(2+\sqrt{3}\right)+i\left\{\dfrac{\pi}{2}+2n\pi\right\}\right]\,\,\,\,\,\,or\,\,\,\,\,\,\theta=-i\cdot\left[\ln\left(2-\sqrt{3}\right)+i\left\{\dfrac{\pi}{2}+2m\pi\right\}\right]}$where, m, n are integers

$\sf{\implies\,\theta=-i\,\ln\left(2+\sqrt{3}\right)-i^2\left\{\dfrac{\pi}{2}+2n\pi\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\theta=-i\,\ln\left(2-\sqrt{3}\right)-i^2\left\{\dfrac{\pi}{2}+2m\pi\right\}}$

$\sf{\implies\,\theta=-i\,\ln\left(2+\sqrt{3}\right)+\left\{\dfrac{\pi}{2}+2n\pi\right\}\,\,\,\,\,\,or\,\,\,\,\,\,\theta=-i\,\ln\left(2-\sqrt{3}\right)+\left\{\dfrac{\pi}{2}+2m\pi\right\}}$

$\sf{\implies\,\theta=\dfrac{\pi}{2}+2n\pi-i\,\ln\left(2+\sqrt{3}\right)\,\,\,\,\,\,or\,\,\,\,\,\,\theta=\dfrac{\pi}{2}+2m\pi-i\,\ln\left(2-\sqrt{3}\right)}$

$\sf{\implies\,\theta=(4n+1)\dfrac{\pi}{2}-i\,\ln\left(2+\sqrt{3}\right)\,\,\,\,\,\,or\,\,\,\,\,\,\theta=(4m+1)\dfrac{\pi}{2}-i\,\ln\left(2-\sqrt{3}\right)}$