Question

Simplify....Cos^-1(3/5cosx+4/5sinx)

Answer 1

using rule of inverse trigonometric function..

we can solve this

see attachment
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hope it will help u

Answer 2

The value of $\bold{\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)\ is\ x-\tan ^{-1}\left(\frac{4}{3}\right)}$

Solution:

Let $\cos \theta=\frac{3}{5}$ ; $r \sin \theta=\frac{4}{5} \dots \ldots \ldots \ldots(i)$

Then squaring and adding both the $r \cos \theta \text { and } r \sin \theta$

$(r \cos \theta)^{2}+(r \sin \theta)^{2}=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}$

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{9}{25}+\frac{10}{25}$

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{25}{25}$

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1$

As we know that $\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1,$

Then, $r^{2}=1$

$r=\sqrt{1} \ldots \ldots \ldots(\text { ii) }$

By Dividing  $r \sin \theta \text { and } \cos \theta$ we can get,  

$\frac{r \sin \theta}{r \cos \theta}=\frac{4}{3}$

$\frac{\sin \theta}{\cos \theta}=\tan \theta$

Therefore, $\tan \theta=\frac{4}{3}$

$\theta=\tan ^{-1}\left(\frac{4}{3}\right) \dots \ldots \ldots(i i i)$

From the equations (i), (ii), (iii)  

$\cos \theta=\frac{3}{5} ; \sin \theta=\frac{4}{5^{\prime}} \theta=\tan ^{-1}\left(\frac{4}{3}\right)$

$\Rightarrow \cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)$

$=\cos ^{-1}(\cos \theta \cos x+\sin \theta \sin x)$

$=\cos ^{-1}[\cos (x-\theta)]$

$=x-\theta$

Substitute the value $\theta=\tan ^{-1}\left(\frac{4}{3}\right)$

$=x-\tan ^{-1}\left(\frac{4}{3}\right)$