Question

How do you prove that `1/2 sin^(-1) (4/5) = tan^(-1) (1/2)` ?

Answer 1

See the Explanation.

Explanation:

Let, `sin^-1(4/5)=alpha, and, tan^-1(1/2)=beta.`

Then, we have to prove that, `1/2alpha=beta, or, alpha=2beta.`

Now, `sin^-1(4/5)=alpha rArr sinalpha=4/5, -pi/2lealphalepi/2.`

But, `sinalpha gt 0 rArr 0ltalphaltpi/2.`

Similarly, `tan^-1(1/2)=beta rArr tanbeta=1/2, 0ltbetaltpi/2.`

In fact, `tanbeta=1/2lt1=tan(pi/4), i.e., tanbeta lt tan(pi/4)...(1)`

Recall that, `tan` fun. is `uarr` in `(0,pi/2):. by,(1), 0ltbetaltpi/4.`

Since, `sec^2beta=1+tan^2beta=1+(1/2)^2=5/4`

`:. secbeta=+sqrt5/2, because, 0ltbetaltpi/4ltpi/2.`

`:. cosbeta=2/sqrt5, and, as, sinbeta=tanbeta*cosbeta,`

`sinbeta=1/sqrt5.`

`:. sin2beta=2sinbetacosbeta=2*1/sqrt5*2/sqrt5=4/5=sinalpha.`

`:. 0 lt beta lt pi/4 rArr 0 lt beta lt pi/2.`

`"Thus, "sinalpha=sin2beta," where, o"lt alpha, beta lt pi/2.`

`:. alpha=2beta,` as desired.

Enjoy Maths.!

Answer 2

`sin(2tan^-1(1/2))`

`=(2tan(tan^-1(1/2)))/(1-tan(tan^-1(1/2)))`

`=(2xx1/2)/(1+(1/2)^2)`

`=1/(1+1/4)`

`=1/(5/4)`

`=4/5`

So

`sin(2tan^-1(1/2))=4/5`

`=>2tan^-1(1/2)=sin^-1(4/5)`

`=>tan^-1(1/2)=1/2sin^-1(4/5)`

Proved

Answer 3

See below.

Explanation:

Think that

`arctan(a/b)=arcsin(a/sqrt(a^2+b^2))`

and

`arcsin(a/b)=arccos(sqrt(b^2-a^2)/b)`

then

`arctan(1/2)=arcsin(1/sqrt(2^2+1))=arcsin(1/sqrt5)`

so

`4/5 = sin(2arcsin(1/sqrt5))`

but `sin(2phi)=2sin(phi)cos(phi)` so

`4/5=2sin(arcsin(1/sqrt5))cos(arcsin(1/sqrt5))`

and

`4/5=2/sqrt(5)cos(arccos(2/sqrt(5)))=(2/sqrt(5))^2`

Answer 4

Use: `(2+i)^2 = 3+4i` and de Moivre's formula...

Explanation:

Note that:

`(2+i)^2 = 3+4i`

Let `alpha` be the most acute angle in a right angled triangle with sides: `1`, `2`, `sqrt(5)`.

Then `tan alpha = 1/2`

By de Moivre's formula, we find:

`3+4i = (2+i)^2`

`color(white)(3+4i) = (sqrt(5)(cos alpha + i sin alpha))^2`

`color(white)(3+4i) = 5(cos 2alpha + i sin 2alpha)`

So equating imaginary parts we find:

`4 = 5 sin 2 alpha`

So:

`4/5 = sin 2 alpha`

So:

`sin^(-1) (4/5) = 2 alpha`

So:

`1/2 sin^(-1) (4/5) = alpha = tan^(-1) (1/2)`