Prove: sin A + sin B + sin C >2 if A+B+C=180 A<90 B<90 C<90?

2 Answers | Add Yours

innainnainna's profile pic

innainnainna | Student | eNotes Newbie

Posted on

Is it proof?

 

You just verify the inquality for certain values.  What if it isn't correct for another set of values?

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the constraint:

A+B = 180-C

We'll apply sine function both sides:

sin(A+B) = sin(180-C) = sin C

We'll divide by 2 the constraint:

(A+B)/2 = (180-C)/2

We'll apply sine function both sides:

sin (A+B)/2 = sin (180-C)/2 = sin (90 -C/2) = cos C/2

Now, we'll manage the left side of the inequality, replacing sin C by sin(A+B):

sin A + sin B + sin(A+B) > 2

We'll transform into a product the sum sin A + sin B:

2 sin [(A+B)/2]* cos [(A-B)/2] + sin(A+B) > 2

We'll use the double angle identity for sin(A+B):

sin(A+B) = 2 sin [(A+B)/2]* cos [(A+B)/2]

2 sin [(A+B)/2]* cos [(A-B)/2] + 2 sin [(A+B)/2]* cos [(A+B)/2]>2

2 sin [(A+B)/2]* {cos [(A-B)/2] +cos [(A+B)/2]} > 2

sin [(A+B)/2]* {cos [(A-B)/2] +cos [(A+B)/2]} > 1

We'll transform into a product the sum cos [(A-B)/2] +cos [(A+B)/2]:

2*sin [(A+B)/2]*cos [(A-B+A+B)/4]*cos[(A-B-A-B)/4] > 1

2*sin [(A+B)/2]*cos [(2A)/4]*cos[(-2B)/4] > 1

We'll replace sin [(A+B)/2] by cos C/2, we'll simplify inside brackets and we'll cancel the minus sign, since the cosine function is even.

cos (A/2) *cos (B/2) *cos (C/2)> 1/2

Since the values of the angles are smaller than 90, then all the angles are located in the 1st quadrant, where the values of the function cosine are positive.(the direction of the inequality remains unchanged).

We'll consider the values of the angle A = 60, B=45 => C = 75, for the sum of the angles to be of 180 degrees.

cos 60/2* cos 45/2* cos 75/2 > 1/2

We'll use the half angle identity:

cos 45/2 = sqrt [(1+cos 45)/2]

cos75/2 = sqrt [(1+cos 75)/2]

cos 60/2*sqrt [(1+cos 45)/2]*sqrt [(1+cos 75)/2] > 1/2

We'll raise to square to eliminate the square root:

3/4*[(1 + sqrt2/2)/2]*{[1+sqrt2/2*(sqrt3/2 - 1/2)]/2} > 1/4

Computing the left side, we'll get:

(12 + 3sqrt6 + 6sqrt2 - 3sqrt3 - 3sqrt3 - 3)/4 > 4

25.74/4 > 4

6.435 > 4

Therefore, the identity sin A + sin B + sin C >2 is verified, under the given constraints.

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question