Question

find the value of cos^2 73 + cos^2 47 + cos47xcos73
please answer.
I'm stuck

Answer 1

1/2[2cos^(47)+2cos^(73)+2cos(47)cos(73)]

=1/2[1+cos(94)+1+cos(146)+cos(120)+cos(26)]

=1/2[2+cos(94)+cos(146)-(1/2)+cos(26)]

=1/2[3/2+{cos(94)+cos(26)}+cos(146)]

=1/2[3/2+2cos(120)cos(34)+cos(146)]

=1/2[3/2+2(1/2)cos(34)+cos(146)]

=1/2[3/2+cos(34)+cos(146)]

=1/2[3/2+2cos(90)cos(56)]

=1/2[3/2+2(0).cos(56)]

=1/2[3/2+0]

=3/4 Ans.

Answer 2

Correct question :-

Find the value of :-

$\sf{cos^2 73^\circ + cos^2 47 ^\circ +cos 73^\circ cos 47^\circ}$

$\rule{190}6$

Solution :-

$\sf{cos^2 73^\circ + cos^2 47 ^\circ +cos 73^\circ cos 47^\circ}\\\\\displaystyle{\sf{= \frac{1}{2}(2cos^2 73+2 cos^2 47+2 cos73.cos47) }}\\\\\bullet\ \sf{As,\ 2cos^2\theta=cos2\theta+1}\\\\\bullet\ \sf{As,\ cosA+cosB=2cos(\frac{A+B}{2} )cos(\frac{A-B}{2} )}\\\\\displaystyle{\sf{= \frac{1}{2} (1+cos\ 147+1+cos\ 94+cos\ 120+cos\ 26)}}\\\\\displaystyle{\sf{= \frac{1}{2} (2-\frac{1}{2}+cos\ 94+cos\ 26+cos\ 146 )}}$

$\displaystyle{\sf{= \frac{1}{2}[\frac{3}{2}+2cos(\frac{120}{2})cos(\frac{68}{2} )+cos\ 146] }}$

$\displaystyle{\sf{= \frac{1}{2} (\frac{3}{2}+2 \times \frac{1}{2}\ cos\ 34+cos\ 146)}}\\\\\displaystyle{\sf{= \frac{1}{2} (\frac{3}{2}+cos\ 34+cos\ 146)}}\\\\\displaystyle{\sf{= \frac{1}{2} [\frac{3}{2}+2cos(\frac{180}{2}cos( \frac{112}{2})]}}\\\\\displaystyle{\sf{= \frac{1}{2} (\frac{3}{2}+2cos\ 90.cos\ 56)}}\\\\\displaystyle{\sf{= \frac{1}{2} (\frac{3}{2}+2\times0\times cos\ 56)}}\\\\\displaystyle{\sf{= \frac{1}{2}(\frac{3}{2} +0) }}\\\\\boxed{\boxed{\displaystyle{\sf{= \frac{3}{4} }}}}$