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If #sinx=4/5# what is the value of #cosx#?

1 Answer

Shwetank Mauria

Feb 16, 2016

#cosx=+(3/5) or -(3/5)#

Explanation:

Recall trigonometric identity #sin^2x+cos^2x=1#. Hence #cos^2x=1-sin^2x# or

#cosx=sqrt(1-sin^2x)=sqrt(1-(4/5)^2)=sqrt(1-16/25)=sqrt(9/25)#

i.e. #cosx=+(3/5) or -(3/5)#

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