How do you verify the identity #cos((5pi)/4-x)=-sqrt2/2(cosx+sinx)#?

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Answer 1 · 2017-02-06T08:53:17

Please see below.

We will use the identities #cos(pi+theta)=-costheta# and #cos(A-B)=cosAcosB+sinAsinB#

Hence, #cos((5pi)/4-x)#

= #cos(pi+(pi/4-x))#

= #-cos(pi/4-x)#

= #-[cos(pi/4)cosx+sin(pi/4)sinx]#

= #-[sqrt2/2cosx+sqrt2/2sinx]#

= #-sqrt2/2(cosx+sinx)#