What is the value of cos35°-cos85°+cos155°
A)0
B)1/√3
C)1/√2
D)1
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Answered by
4
Hi Friend,
Here is your answer,
cos 35°- cos85°+cos155°
=> 2cos{(35°+85°)/2]cos[(35°-85°)/2]+cos(180°-25°)
=>[cosC+cosD=2cos[(C+D)/2]cos[(C-D)/2]
=>2cos(120°/2)cos(50°/2)+cos[(90°×2)-25°]
=>2cos60°cos25°+(-cos25°)
=>2×(1/2)cos25°-cos25°
= 0 is the correct answer.
Hope it helps you!
Here is your answer,
cos 35°- cos85°+cos155°
=> 2cos{(35°+85°)/2]cos[(35°-85°)/2]+cos(180°-25°)
=>[cosC+cosD=2cos[(C+D)/2]cos[(C-D)/2]
=>2cos(120°/2)cos(50°/2)+cos[(90°×2)-25°]
=>2cos60°cos25°+(-cos25°)
=>2×(1/2)cos25°-cos25°
= 0 is the correct answer.
Hope it helps you!
mark as brainliest answer!! thanks
Wt u gave is right only but I asked cos 35-cos 85not cos 35+cos85
oh sorry!! very sorry
No problem
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just + - difference is there
Plzz send
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ok i will definitely !
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Answered by
0
cos 35°+cos85°tcos155°
= 2cos{(35°+85°)/2}cos{(35°-85°)/ =
2}+cos(180°-259)
[cosC+cosB=2 cos((C+D)/2}cos((C-D)/2}]
=2cos(120/2)cos(50°/2)+cos((90°x2)-25°)
=2cos60°cos 25°+(-cos 25°) =2x(1/2)cos 25°-cos 25°
=O (Proved)
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