Question

Heya users ☺

Here's a question for you

If x and y are acute angles such that cos x=13/14 and cos y=1/7, prove that (x-y) = -π/3

✳COMPLETE SOLUTION REQUIRED✳

✖PLEASE NO SPAM ANSWERS✖

Answer 1

Given cosx = 13/14

we know that sin^2x + cos^2x = 1

= > sin^2x = 1 - cos^2x

                 = 1 - (13/14)^2

                 = 1 - 169/196

                 = 27/196


$sinx = \frac{ \sqrt{27} }{14}$


Now,

sin^2y = 1 - cos^2y

            = 1 - (1/7)^2

            = 1 - 1/49

            = 48/49.


$siny = \frac{ \sqrt{48} }{7}$


Now,

We know that cos(x - y) = cosxcosy + sinxsiny

$= \ \textgreater \ \frac{13}{14} * \frac{1}{7} + \frac{ \sqrt{27} }{14} * \frac{ \sqrt{48} }{7}$

$= \ \textgreater \ cos(x - y) = \frac{13}{98} + \frac{18}{49}$

$= \ \textgreater \ cos(x - y) = \frac{49}{98}$

$= \ \textgreater \ cos(x - y) = \frac{1}{2}$

= > x - y = pi/3.


Hope this helps!

Answer 2

Hi,

Please see the attached file!



Thanks