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# some trig identities p2 watch

1. heinman p.81 rev ex1
qu 16:
prove (cosx+cosy)^2+(sinx+siny)^2=4cos ^2([x-y]/2)

i can get as far as 2+2[cos(x-y)]
starting from LHS and opening up brackets, getting rid of sin^2 x+ cos^2 x for 1 and same with sin^2 y + cos^2 y for 1 and then finally using cos(a-b) rule but now im stuck....
heinman p.81 rev ex1
qu 16:
prove (cosx+cosy)^2+(sinx+siny)^2=4cos ^2([x-y]/2)

i can get as far as 2+2[cos(x-y)]
starting from LHS and opening up brackets, getting rid of sin^2 x+ cos^2 x for 1 and same with sin^2 y + cos^2 y for 1 and then finally using cos(a-b) rule but now im stuck....
use these formula:
cosA+cosB=2cos((A+B)/2)cos((A-B)/2)
sinA+sinB=2sin((A+B)/2)cos((A-B)/2)
the result should fall out nicely.
heinman p.81 rev ex1
qu 16:
prove (cosx+cosy)^2+(sinx+siny)^2=4cos ^2([x-y]/2)

i can get as far as 2+2[cos(x-y)]
starting from LHS and opening up brackets, getting rid of sin^2 x+ cos^2 x for 1 and same with sin^2 y + cos^2 y for 1 and then finally using cos(a-b) rule but now im stuck....
Or u still could stick with ur method and note that cos2A=2cos²A-1.
heinman p.81 rev ex1
qu 16:
prove (cosx+cosy)^2+(sinx+siny)^2=4cos ^2([x-y]/2)

i can get as far as 2+2[cos(x-y)]
starting from LHS and opening up brackets, getting rid of sin^2 x+ cos^2 x for 1 and same with sin^2 y + cos^2 y for 1 and then finally using cos(a-b) rule but now im stuck....
let x=a+b
let y=a-b

(cosacosb-sinasinb+cosacosb+sinasinb)^2+(s inacosb+cosasinb+sinacosb-cosasinb)^2=4(cos^2acos^2b+sin^2 acos^2b)
=4cos^2b(cos^2a+sin^2a) [and sin^2x+cos^2x=1
=4cos^2b [and as 2b=x-y]
=4cos^2((x-y)/2)

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Updated: June 20, 2004
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