2 sin a + 4 sin b + 3 sin c = ....

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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By definition, `sin alpha in [-1,1]` , hence, you need to consider the following three inequalities, such that:

`{(-1 <= sin a <= 1),(-1 <= sin b <= 1),(-1 <= sin c <= 1):}`

`{(-1*2 <= 2sin a <= 1*2),(-1*4 <= 4*sin b <= 1*4),(-1*3 <= 3sin c <= 1*3):}`

Adding the inequalities, yields:

`-2 - 4 - 3<= 2sin a + 4sin b + 3sin c <= 2 + 4 + 3`

`-9 <= 2sin a + 4sin b + 3sin c <= 9`

Hence, evaluating the possible values of the following summation, `2sin a + 4sin b + 3sin c` , yields` -9 <= 2sin a + ` `4sin b + 3sin c <= 9.`

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

`|sin(alpha)|<=1`  for all  `alpha` .

so.

`2|sin(a)|<=2`

`4|sin(b)|<=4`

`3|sin(c)|<=3`

` ` for all angle ,a,b,and c.

adding above inequalitie we have

`2|sin(a)|+4|sin(b)|+3|sin(c)|<=2+3+4`

`2|sin(a)|+4|sin(b)|+3|sin(c)|<=9`

if 0 <a,b c <90 then

`2sin(a)+4sin(b)+3sin(c)<=9`

 

There all six possible combination in limits change.

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