Question

if cos x/2=12/13 and x lies in Quadrant I, find the values of
(i) sin x,
(ü) cos x,
(iii) cotx.​

Answer 1

Answer:

$sinx=\frac{120}{169}$

$cosx=\frac{119}{169}$

$cotx=\frac{119}{120}$

Step-by-step explanation:

Formula used:

$sin^2{A}+cos^2{A}=1$

$cosA=2\:cos^2\frac{A}{2}-1$

Given:

$cos\frac{x}{2}=\frac{12}{13}$

$cos^2\frac{x}{2}+sin^2\frac{x}{2}=1$

$\frac{144}{169}+sin^2\frac{x}{2}=1$

$sin^2\frac{x}{2}=1-\frac{144}{169}$

$sin^2\frac{x}{2}=\frac{169-144}{169}$

$sin^2\frac{x}{2}=\frac{25}{169}$

$sin\frac{x}{2}=\frac{5}{13}$          ( x/2 lies in I quadrant)

$cosx=2\:cos^2\frac{x}{2}-1$

$cosx=2(\frac{12}{13})^2-1$

$cosx=2(\frac{144}{169})-1$

$cosx=\frac{288}{169}-1$

$cosx=\frac{288-169}{169}$

$cosx=\frac{119}{169}$

$sinx=2\:sin\frac{x}{2}\:cos\frac{x}{2}$

$sinx=2(\frac{5}{13})(\frac{12}{13})$

$sinx=\frac{120}{169}$

$cotx=\frac{cosx}{sinx}$

$cotx=\frac{\frac{119}{169}}{\frac{120}{169}}$

$cotx=\frac{119}{120}$

Answer 2

See the attachment alls are mentioned there .

hope its help u . . . . .