Question

show that sin( log i^i) = -1

Answer 1

Answer:

show that sin( log eii) = 1

Answer 2

Answer:

Proof is given below.

Step-by-step explanation:

We need to show that sin(log $i^{i}$) = -1

We know that $e^{ix}$ = cos x + i sin x

                            = cis x

Putting x = $\frac{\pi}{2}$

=> $e^{i\frac{\pi}{2}}$ = cos  $\frac{\pi}{2}$  + i sin  $\frac{\pi}{2}$

=>  $e^{i\frac{\pi}{2}}$ =0 + i *1

=> $e^{i\frac{\pi}{2}}$ = i

As in this question, we need $i^{i}$ ,

therefore raising the above to the power i

=>  $e^{i\frac{\pi}{2}*i}$ = $i^{i}$

=>  $e^{i^{2}\frac{\pi}{2}}$ = $i^{i}$

=>  $e^{-1\frac{\pi}{2}}$ = $i^{i}$

=>  $e^{-\frac{\pi}{2}}$ = $i^{i}$

Taking log both sides,

=> log (  $e^{-\frac{\pi}{2}}$ ) = log ( $i^{i}$ )

=>  $\frac{-\pi}{2}$ = log ( $i^{i}$ )

As in this question, we need sin(log $i^{i}$),

therefore taking sin both sides,

=> sin( $\frac{-\pi}{2}$ ) = sin( log ( $i^{i}$ ))

=> -1 = sin( log $i^{i}$ )

Hence, the result.