If cosA+cosB+cosC+cosD+cosE = 5 then find sinA+sinB+sinC +sinD + sinE ?
Answers
Answer:
0.
Step-by-step explanation:
Using the basic properties of trigonometric ratios:
cosine of any angle can't be greater than 1.
Here,
We are provided with cosines of 5 angles( which may or may not be equal ).
Maximum numeric value of cosine of any angle can't exceed 1, so if there are 5 cosines and when they are added they are providing 5, which means all of them are having the maximum value a cosine can hold.
Anyhow, if one say cosine of A or B or C or D or E or two/three of them is lesser than 1, then these cosines will never add upto 5.
Thus,
cosA = cosB = cosC = cosD = cosE = 1.
And, we can say that these angles are equal to 0 or periodic at 2π, ( so any angle periodic to 2π ).
Thus,
cos^2 A + sin^2 A = 1
⇒ 1 + sin^2 A = 1 ⇒ sinA = 0,
In the same manner, sinA = sinB = sinC = sinD = sinE = 0.
⇒ sinA + sinB + sinC + sinD + sinE = 0.
Answer:
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