X=cos55°,y=cos65°,z=cos175°, find xy+yz+xz
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38
Given x = cos 55, y = cos 65, z = cos 175.
We need to find xy, yz, zx.
= > cos 55 * cos 65 + cos 65 * cos 175 + cos 175 * cos 55
We know that 2cosAcosB = cos(A + B) + cos(A-B).
= > (1/2)(cos(55 + 65) + cos(65 - 55)) + (1/2)(cos(65 + 175) + cos(175 - 65)) + (1/2)(cos(175 + 55)cos(175 - 55))
= > (1/2)(cos120+cos10) + (1/2)(cos240+cos110) + (1/2)(cos230 + cos120)
= > (1/2)(cos120) + (1/2)cos(10) + (1/2)cos240 + (1/2)cos(110) + (1/2)cos(230) + (1/2)cos120
= > (1/2)cos(180 - 60) + (1/2)cos(10) + (1/2)cos(180 + 60) + (1/2)cos(110) + (1/2)(cos230) - 1/2(cos(180 - 60))
= > (1/2)(-cos60) + (1/2)cos(10) + (1/2)cos(-cos60) + (1/2)cos110 + (1/2)cos230 - 1/2(-cos60)
= > (1/2)(-1/2) + (1/2)cos(10) + (1/2)(-1/2) + (1/2)cos(110) + (1/2)cos(230) - (1/2)(-1/2)
= > -1/4 + 1/2cos(10) - 1/4 + (1/2)cos(110) + (1/2)cos230 - 1/4
= > -(3/4) + (1/2)(cos10 + cos110 + cos230)
Hope this helps!
We need to find xy, yz, zx.
= > cos 55 * cos 65 + cos 65 * cos 175 + cos 175 * cos 55
We know that 2cosAcosB = cos(A + B) + cos(A-B).
= > (1/2)(cos(55 + 65) + cos(65 - 55)) + (1/2)(cos(65 + 175) + cos(175 - 65)) + (1/2)(cos(175 + 55)cos(175 - 55))
= > (1/2)(cos120+cos10) + (1/2)(cos240+cos110) + (1/2)(cos230 + cos120)
= > (1/2)(cos120) + (1/2)cos(10) + (1/2)cos240 + (1/2)cos(110) + (1/2)cos(230) + (1/2)cos120
= > (1/2)cos(180 - 60) + (1/2)cos(10) + (1/2)cos(180 + 60) + (1/2)cos(110) + (1/2)(cos230) - 1/2(cos(180 - 60))
= > (1/2)(-cos60) + (1/2)cos(10) + (1/2)cos(-cos60) + (1/2)cos110 + (1/2)cos230 - 1/2(-cos60)
= > (1/2)(-1/2) + (1/2)cos(10) + (1/2)(-1/2) + (1/2)cos(110) + (1/2)cos(230) - (1/2)(-1/2)
= > -1/4 + 1/2cos(10) - 1/4 + (1/2)cos(110) + (1/2)cos230 - 1/4
= > -(3/4) + (1/2)(cos10 + cos110 + cos230)
Hope this helps!
siddhartharao77:
:-)
Answered by
6
Answer:
-3/4
Step-by-step explanation:
Hpe it helps
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