need an aswer quich ok thiers two the find sin(2pi/3 -pi/4) using sum and difference identites and then tan 5pi/12 using the same identites
You have a formula for
Sin(A-B)
Use that on the first one.
Now on the second, recall that
5PI/12= 3PI/12 + 2PI/12=PI/4 +PI/6
You now can use the tan (A+B) formula, as you know the tan(PI/4) and tan(PI/6)
To find the value of sin(2π/3 - π/4) and tan(5π/12) using sum and difference identities, let's break down the steps:
1. Recall the sum and difference identities for sine and cosine:
a) sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B)
b) cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
2. For sin(2π/3 - π/4):
a) Convert 2π/3 and π/4 to their respective sine and cosine values:
- We know that sin(π/6) = 1/2 and cos(π/6) = √3/2.
b) Use the sum identity for sine:
- sin(2π/3 - π/4) = sin(2π/3)cos(π/4) - cos(2π/3)sin(π/4)
c) Substitute the values from step 2a:
- sin(2π/3) = sin(π/6 + π/2) = sin(π/6)cos(π/2) + cos(π/6)sin(π/2)
= (1/2)(0) + (√3/2)(1) = √3/2.
- cos(2π/3) = cos(π/6 + π/2) = cos(π/6)cos(π/2) - sin(π/6)sin(π/2)
= (√3/2)(0) - (1/2)(1) = -1/2.
d) Substitute the values from step 2c into the sum identity:
- sin(2π/3 - π/4) = (√3/2)(1/√2) - (-1/2)(1/√2)
= (√3/2√2) + (1/2√2)
= (√3 + 1)/(2√2)
= (√6 + √2)/(4).
3. For tan(5π/12):
a) Convert 5π/12 to its respective sine and cosine values:
- We know that sin(π/6) = 1/2 and cos(π/6) = √3/2.
b) Use the difference identity for tangent:
- tan(5π/12) = [tan(π/3) - tan(π/6)] / [1 + tan(π/3)tan(π/6)]
c) Substitute the values from step 3a:
- tan(π/3) = (sin(π/3)) / (cos(π/3)) = (√3/2) / (1/2) = √3.
- tan(π/6) = (sin(π/6)) / (cos(π/6)) = (1/2) / (√3/2) = 1/√3.
d) Substitute the values from step 3c into the difference identity:
- tan(5π/12) = [√3 - (1/√3)] / [1 + (√3)(1/√3)]
= (√3 - 1)/(1 + 1)
= (√3 - 1)/2.
Therefore, the value of sin(2π/3 - π/4) is (√6 + √2)/(4), and the value of tan(5π/12) is (√3 - 1)/2.