Question #8ab7c

Question

Question #8ab7c

1 Answer

Impact of this question

1305 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-05T05:04:22

$cos (x - 5 \frac{π}{6}) = \frac{1}{2} {\sqrt{1 - {cos}^{2} x} - \sqrt{3} cos x}$ $cos(x-5pi/6)=1/2{sqrt(1-cos^2x)-sqrt3cosx}$ .

Explanation:

Prerequisites $(1) : cos (A - B) = cos A cos B + sin A sin B$ $(1): cos(A-B)=cosAcosB+sinAsinB$ ,

$(2) (i) : cos (π - θ) = - cos θ, (2) (i i) : sin (π - θ) = sin θ$ $(2)(i): cos(pi-theta)=-costheta, (2)(ii) : sin(pi-theta)=sintheta$ ,

$(3) (i) : cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}, (3) (i i) sin (\frac{π}{6}) = \frac{1}{2}$ $(3)(i): cos(pi/6)=sqrt3/2, (3)(ii) sin(pi/6)=1/2$ ,

$(4) : sin θ = \sqrt{1 - {cos}^{2} θ}$ $(4): sintheta=sqrt(1-cos^2theta)$ .

Now, $cos (x - 5 \frac{π}{6}) = cos x cos (5 \frac{π}{6}) + sin x sin (5 \frac{π}{6})$ $cos(x-5pi/6)=cosxcos(5pi/6)+sinxsin(5pi/6)$ ,

$= cos x cos (π - \frac{π}{6}) + sin x sin (π - \frac{π}{6})$ $=cosxcos(pi-pi/6)+sinxsin(pi-pi/6)$ ,

$= (- cos (\frac{π}{6})) cos x + sin (\frac{π}{6}) sin x$ $=(-cos(pi/6))cosx+sin(pi/6)sinx$ ,

$= - \frac{\sqrt{3}}{2} cos x + \frac{1}{2} sin x$ $=-sqrt3/2cosx+1/2sinx$ ,

$= \frac{1}{2} (sin x - \sqrt{3} cos x)$ $=1/2(sinx-sqrt3cosx)$ .

$\Rightarrow cos (x - 5 \frac{π}{6}) = \frac{1}{2} {\sqrt{1 - {cos}^{2} x} - \sqrt{3} cos x},$ $rArr cos(x-5pi/6)=1/2{sqrt(1-cos^2x)-sqrt3cosx},$ is the

desired expression!

Science

Math

Humanities

... and beyond

Question #8ab7c

1 Answer

Explanation:

Related questions

Impact of this question