cos-1 [ 3/5 cosx + 4/5sinx ] find its value I can ' t - Maths - Inverse Trigonometric Functions

Question

cos -1 [ 3/5 cosx +
4/5sinx ] find its value?

I can ' t understand this question please explain it briefly
?

Aakash Sharma · Accepted Answer

$Let \frac{3}{5} = r \cos θ and \frac{4}{5} = r \sin θ ...... (1)$

Squaring and adding we get

$(r \cos θ)^{2} + (r \sin θ)^{2} = (\frac{3}{5})^{2} + (\frac{4}{5})^{2} \Rightarrow r^{2} (\cos^{2} θ + \sin^{2} θ) = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1 \Rightarrow r = \sqrt{1} = 1 ..... (2)$

dividing them we get

$\frac{r \sin θ}{r \cos θ} = \frac{4}{3} \Rightarrow \tan θ = \frac{4}{3} \Rightarrow θ = \tan^{- 1} \frac{4}{3} ..... (3)$

From (1), (2) and (3) we get

$\cos θ = \frac{3}{5} and sin θ = \frac{4}{5} where θ = \tan^{- 1} \frac{4}{3}$

$∴ \cos^{- 1} (\frac{3}{5} \cos x + \frac{4}{5} \sin x) = \cos^{- 1} (\cos θ \cos x + \sin θ \sin x) = \cos^{- 1} [\cos (x - θ)] = x - θ = x - \tan^{- 1} \frac{4}{3}$

cos -1 [ 3/5 cosx + 4/5sinx ] find its value? I can ' t understand this question . please explain it briefly ?

cos^-1 [ 3/5 cosx + 4/5sinx ] find its value?

I can ' t understand this question . please explain it briefly ?