How do you find the exact value of #cos62# using the sum and difference, double angle or half angle formulas?

1 Answer
Jan 8, 2018

#cos(62^@)=cos((7pi)/12)=(sqrt(2)-sqrt(6))/4#

Explanation:

Best way to start is to convert the angle measure from degrees to radians. A quick refresher on how to do this;

Radians = Degrees #xxpi/180#

Radians = #62xxpi/180=(62pi)/180=(7pi)/12#

#:.cos(62^@)=cos((7pi)/12)#

#cos((7pi)/12)=cos(pi/3+pi/4)#

The trigonometric identity that we are going to use is the angle sum identity for cosine;

#color(red)(cos(A+B)=cos(A)cos(B)-sin(A)sin(B))#

Entering in the appropriate angles;

#cos(pi/3+pi/4)=cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)#

Evaluating the trigonometric ratios yields;

#cos(pi/3+pi/4)=(1/2)(sqrt(2)/2)-(sqrt(3)/2)(sqrt(2)/2)#

Hence,

#cos(pi/3+pi/4)=sqrt(2)/4-sqrt(6)/4=(sqrt(2)-sqrt(6))/4#

I hope that is helpful,

If you like I have a youtube channel with heaps of videos on problems like this, you should check it out:

https://www.youtube.com/haroldwalden

:)