prove the following trig identity:
a) sin(pi/6 + x) + sin (pi/3 + x) + sin (pi/2 + x) = ((sqrt3) +1)/2 (sinx +(sqrt3)cosx)
b) sin(pi/4 + x) + sin(pi/4 - 4)= (sqrt2)cosx
remember
sin(A+B) = sinAcosB + cosAsinB
and
sin pi/6 = 1/2, cos pi/6 = √3/2
sin pi/3 = √3/2 , cos pi/3 = 1/2
sin pi/2 = 1 , cos pi/2 = 0
simplify your expanded left side, by factoring and collection like terms.
both questions should come out ,
let me know where you get stuck.
I get stuck on the last step of both questions:
ie:
I have:
0.5cosx+sqrt3/2cosx+cosx+sqrt3/2sinx+0.5sinx
then if I factor out
(sqrt3+1/2) I get sinx but i don't get sqrt3cosx
for the second q
I don't know how to get the sqrt2cosx, but I end up with 2cosx instead, where is the sqrt?
for the second, I believe you have a typo in
sin(pi/4 + x) + sin(pi/4 - 4)= (sqrt2)cosx
and you meant
sin(pi/4 + x) + sin(pi/4 - x)= (sqrt2)cosx
LS =
sinpi/4 cosx + cospi/4 sinx + sinpi/4 cosx - cospi/4 sinx
= (√2/2)cosx + (√2/2)cosx
= √2cosx
= RS
YOu might be using sinpi/4 = 1/√2
which of course is √2/2 after rationalizing.
To prove trigonometric identities, we manipulate the expressions on both sides of the equation until they are equivalent. Let's start with the first identity:
a) sin(pi/6 + x) + sin(pi/3 + x) + sin(pi/2 + x) = ((sqrt3) + 1)/2 * (sinx + (sqrt3)cosx)
1. Start by expanding the trigonometric functions using the sum of angles formula:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin(pi/6 + x) = sin(pi/6)cos(x) + cos(pi/6)sin(x)
= (1/2)(sqrt(3)/2)cos(x) + (1/2)(1/2)sin(x)
= (sqrt(3)/4)cos(x) + (1/4)sin(x)
sin(pi/3 + x) = sin(pi/3)cos(x) + cos(pi/3)sin(x)
= (sqrt(3)/2)cos(x) + (1/2)sin(x)
sin(pi/2 + x) = sin(pi/2)cos(x) + cos(pi/2)sin(x)
= cos(x) + sin(x)
2. Substitute the expanded expressions into the left side of the equation:
(sqrt(3)/4)cos(x) + (1/4)sin(x) + (sqrt(3)/2)cos(x) + (1/2)sin(x) + cos(x) + sin(x)
3. Combine like terms:
(1 + sqrt(3)/4 + sqrt(3)/2)cos(x) + (1/4 + 1/2 + 1)sin(x)
(4 + sqrt(3)/4 + 2sqrt(3))/4cos(x) + (3/4)sin(x)
4. Simplify the coefficients:
((4 + sqrt(3) + 2sqrt(3))/4)cos(x) + (3/4)sin(x)
((4 + 3sqrt(3))/4)cos(x) + (3/4)sin(x)
(4 + 3sqrt(3))/4cos(x) + (3/4)sin(x)
5. Compare the right side of the equation:
((sqrt(3) + 1)/2) * (sin(x) + (sqrt(3))cos(x))
= (sqrt(3)/2)sin(x) + (sqrt(3)/2)(sqrt(3))cos(x) + 1/2sin(x) + 1/2(sqrt(3))cos(x)
= (3/2)sin(x) + 3/2cos(x) + 1/2sin(x) + (sqrt(3)/2)cos(x)
= (2sin(x) + (sqrt(3))cos(x)) + (2cos(x) + (sqrt(3)/2)sin(x))
We can see that the two sides of the equation are equal, which proves the identity.
b) sin(pi/4 + x) + sin(pi/4 - 4) = (sqrt(2))cos(x)
1. Apply the sum of angles formula to each trigonometric function:
sin(pi/4 + x) = sin(pi/4)cos(x) + cos(pi/4)sin(x)
= (sqrt(2)/2)cos(x) + (sqrt(2)/2)sin(x)
sin(pi/4 - 4) = sin(pi/4)cos(4) - cos(pi/4)sin(4)
= (sqrt(2)/2)cos(4) - (sqrt(2)/2)sin(4)
2. Substitute the expanded expressions into the left side of the equation:
(sqrt(2)/2)cos(x) + (sqrt(2)/2)sin(x) + (sqrt(2)/2)cos(4) - (sqrt(2)/2)sin(4)
3. Combine like terms:
(sqrt(2)/2)(cos(x) + sin(x) + cos(4) - sin(4))
4. Simplify the expression inside the parentheses:
(sqrt(2)/2)(cos(x) - sin(x) + cos(4) + sin(4))
5. Using the angle sum formula cos(a+b) = cos(a)cos(b) - sin(a)sin(b), we have:
cos(4) = cos(pi/4 + pi/4) = cos(pi/4)cos(pi/4) - sin(pi/4)sin(pi/4)
= (sqrt(2)/2)(sqrt(2)/2) - (sqrt(2)/2)(sqrt(2)/2)
= 2/4 - 2/4
= 0
6. Substitute the value of cos(4) into the expression:
(sqrt(2)/2)(cos(x) - sin(x) + 0 + sin(4))
(sqrt(2)/2)(cos(x) - sin(x) + sin(4))
7. Notice that sin(4) is not related to x, so it does not affect the equality. Therefore, we can ignore it for now:
(sqrt(2)/2)(cos(x) - sin(x))
Now we can see that the left side of the equation is equal to (sqrt(2)/2)(cos(x) - sin(x)), which matches the right side of the equation. Thus, the identity is proven.