sin240+sin120+sin150+sin210=?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let S = sin240 + sin120 + sin150 + sin210

We know that:

sin(180 + a) = -sina

sin (180 -a ) = sina

Now, let us rewrite each term:

sin240 = sin(180 + 60 ) = -sin60 = - sqrt3/2

sin120 = sin(180 - 60) = sin60 = sqrt3/2

sin150 = sin(180 - 30) = sin30 = 1/2

sin210 = sin(180 + 30) = -sin30 = -1/2

Now let us add all terms:

S = -sqrt3/2 + sqrt3/2 + 1/2 - 1/2 = 0

Then the sum = 0.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

sin240+sin120+sin150+sin210

sin 240 = sin(180+60) = -sin 60.

sin120 = sin(180-60) = sin60

sin150 = sin(180-30) = sin30

sin210 = sin(180+30) = -sin30

Adding we get:

sin240+sin120+sin150+sin210 = -sin60 +sin60 +sin30-sin30 = 0

So sin 240+sin120+sin150+sin210 = 0

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the value of the sum, we'll apply the formula that transforms the sum of sine functions into a product:

sin a + sin b = 2 sin[(a+b)/2]*cos[(a-b)/2]

We'll combine the first and the second term:

sin 120 + sin 240 = 2 sin[(120+240)/2]*cos[(120-240)/2]

sin 120 + sin 240 = 2 sin[(360)/2]*cos[(-120)/2]

sin 120 + sin 240= 2 sin 180*cos (-60)

But sin 180 = 0, so:

sin 120 + sin 240 = 0

We'll combine the third and the last term:

sin150+sin210 = 2 sin[(150+210)/2]*cos[(150-210)/2]

sin150+sin210 = 2 sin (360/2) *cos (-60/2)

sin150+sin210 = 2 sin 180*cos (-30)

But cos(-30) = cos 30,because the function cosine is an even function.

sin150+sin210 = 2*0*cos 30

sin150+sin210 = 0

So, the value of the sum is:

sin240+sin120+sin150+sin210 = 0+0 = 0

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