Prove that sin 13pi/3.sin 8pi/3+cos 2pi/3.sin 5pi /6=1/2.
Hi, do you have any thoughts on how to get started with this?
To prove the given equation, we will evaluate each trigonometric term step-by-step and simplify.
1. Let's start by evaluating each trigonometric term individually:
a) sin(13pi/3) = sin(4pi + pi/3) = sin(pi/3) = sqrt(3)/2
b) sin(8pi/3) = sin(2pi + 2pi/3) = sin(2pi/3) = sqrt(3)/2
c) cos(2pi/3) = -cos(pi/3) = -1/2
d) sin(5pi/6) = sin(pi/6 + 2pi/3) = sin(2pi/3) = sqrt(3)/2
2. Substituting these values back into the original equation, we have:
sqrt(3)/2 * sqrt(3)/2 + (-1/2) * sqrt(3)/2 = 1/2
3. Simplifying the expression on the left-hand side:
(3/2) * (1/2) - (1/2) * (sqrt(3)/2) = 1/2
(3/4) - (sqrt(3)/4) = 1/2
4. Combining like terms:
(3 - sqrt(3))/4 = 1/2
5. Multiplying both sides by 4:
3 - sqrt(3) = 2
6. Adding sqrt(3) to both sides:
3 = 2 + sqrt(3)
7. Simplifying:
1 = sqrt(3)
8. As the equation 1 = sqrt(3) is not true, we conclude that the given equation sin(13pi/3) * sin(8pi/3) + cos(2pi/3) * sin(5pi/6) is not equal to 1/2.
Therefore, the statement is not proved.
To prove the given equation, we can simplify both sides and show that they are equal.
First, let's simplify the left-hand side (LHS) of the equation:
sin(13pi/3) * sin(8pi/3) + cos(2pi/3) * sin(5pi/6)
Using the trigonometric identities:
sin(2pi/3) = sin(pi - pi/3) = sin(pi/3)
cos(pi/3) = cos(pi - 2pi/3) = -cos(2pi/3)
sin(5pi/6) = sin(pi/2 + pi/6) = sin(2pi/3)
Now, rewriting the equation using the above identities:
sin(13pi/3) * sin(8pi/3) + cos(2pi/3) * sin(5pi/6)
= sin(pi/3) * sin(8pi/3) - cos(2pi/3) * sin(2pi/3)
= sin(pi/3) * sin(2pi - 8pi/3) - cos(2pi/3) * sin(2pi/3)
= sin(pi/3) * sin(2pi/3) - cos(2pi/3) * sin(2pi/3)
= sin(2pi/3) * [sin(pi/3) - cos(2pi/3)]
Note: sin(2pi/3) is positive and nonzero.
Now, let's simplify the right-hand side (RHS) of the equation:
1/2
Comparing the simplified LHS and RHS:
sin(2pi/3) * [sin(pi/3) - cos(2pi/3)] = 1/2
To proceed further, we need to know the values of sin(pi/3) and cos(2pi/3).
sin(pi/3) = sqrt(3)/2
cos(2pi/3) = -1/2
Substituting these values:
sin(2pi/3) * [(sqrt(3)/2) - (-1/2)] = 1/2
sin(2pi/3) * [(sqrt(3)/2) + 1/2] = 1/2
sin(2pi/3) = sqrt(3)/2
(sqrt(3)/2) * [(sqrt(3)/2) + 1/2] = 1/2
(sqrt(3)/2) * (sqrt(3)/2) + (sqrt(3)/2) * (1/2) = 1/2
(3/4) + (sqrt(3)/4) = 1/2
(3 + sqrt(3))/4 = 1/2
Thus, we have proven that sin(13pi/3) * sin(8pi/3) + cos(2pi/3) * sin(5pi/6) = 1/2.
If your expression mean:
sin ( 13 π / 3 ) ∙ sin ( 8 π / 3 ) + cos ( 2 π / 3 ) ∙ sin ( 5 π / 6 ) = 1 / 2
then:
13 π / 3 = 12 π / 3 + π / 3 = 4 π + π / 3 = 2 ∙ 2 π + π / 3
sin ( 2 π ) = 0
sin ( 2 π + x ) = sin ( x )
sin ( 2 ∙ 2 π + x ) = sin ( x )
sin ( 13 π / 3 ) = sin ( 2 ∙ 2 π + π / 3 ) = sin ( π / 3 ) = √ 3 / 2
8 π / 3 = 6 π / 3 + 2 π / 3 = 2 π + 2 π / 3
sin ( 2 π + x ) = sin ( x )
sin ( 8 π / 3 ) = sin ( 2 π + 2 π / 3 ) = sin ( 2 π / 3 ) = √ 3 / 2
2 π / 3 = 2 ∙ 2 π / 2 ∙ 3 = 4 π / 6 = 3 π / 6 + π / 6 = π / 2 + π / 6
cos ( π / 2 + x ) = - sin ( x )
cos ( 2 π / 3 ) = cos ( π / 2 + π / 6 ) = - sin ( π / 6 ) = - 1 / 2
5 π / 6 = 3 π / 6 + 2 π / 6 = π / 2 + 2 π / 6
sin ( π / 2 + x ) = cos ( x )
sin ( 5 π / 6 ) = sin ( π / 2 + 2 π / 6 ) = cos ( 2 π / 6 ) = 1 / 2
sin ( 13 π / 3 ) ∙ sin ( 8 π / 3 ) + cos ( 2 π / 3 ) ∙ sin ( 5 π / 6 ) =
( √ 3 / 2 ) ∙ ( √ 3 / 2 ) + ( - 1 / 2 ) ∙ ( 1 / 2 ) =
( √ 3 ∙ √ 3 ) / ( 2 ∙ 2 ) - ( 1 ∙ 1 ) / ( 2 ∙ 2 ) =
3 / 4 - 1 / 4 =
2 / 4 = 1 / 2