Math, asked by kumarsai, 1 year ago

cos square 45 degree minus sin square 15 degrees

Answers

Answered by goodday
7
cos 45 = 1/ \sqrt{2} sin 45 = 1/ \sqrt{2} sin 30 = 1/2
sin 15 = sin45-sin30
    =1/ \sqrt{2}  -1/2 = 2-tex] \sqrt{2} [/tex]/2tex] \sqrt{2} [/tex]
cos45-sin15 =  2 -tex] \sqrt{2} [/tex]/2tex] \sqrt{2} [/tex]-1/1/ \sqrt{2}  
=  \sqrt{2}  /2/ \sqrt{2}  
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