Question

1) 6sin(pi/12)cos(pi/12) = 2) sin^2(5A)-cos^2(5A) = 3) (2tg2A) / (1-tg^2(2A)) = 4) sin(A/6)cos(A/6) = 5) 20sin^2(4A)cos^2(4A) = 6) sin3Acos3A =

Answer 1

1) (6sin(π/12))*(cos(π/12))=3sinπ/6=3*0.5=1.5

2) sin²(5α)-cos²(5α)= -cos10α

3)(2tg2α)/(1-tg²(2α))=tg4α

4)sin( A/6)cos(A/6)=0.5sin (A/3)

5) 20sin²(4α)cos²(4α)=(5*(2sin(4α))*cos(4α))²=5sin²8α

6)(sin3α)(cos3α)=0.5sin6α

Answer 2

Let's solve each trigonometric expression one by one:

1. 6sin(pi/12)cos(pi/12):

Recall the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)

So, 6sin(π/12)cos(π/12) = 6 * (1/2) * sin(π/6) = 6 * (1/2) * (1/2) = 3/2

Answer: 3/2

2. sin^2(5A) - cos^2(5A):

Recall the Pythagorean identity: sin^2(θ) + cos^2(θ) = 1

sin^2(5A) - cos^2(5A) = (sin^2(5A) + cos^2(5A)) - 2cos^2(5A) = 1 - 2cos^2(5A)

Answer: 1 - 2cos^2(5A)

3. (2tan(2A)) / (1 - tan^2(2A)):

Recall the double angle formula for tangent: tan(2θ) = (2tan(θ)) / (1 - tan^2(θ))

So, (2tan(2A)) / (1 - tan^2(2A)) = tan(4A)

Answer: tan(4A)

4. sin(A/6)cos(A/6):

Recall the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)

sin(A/6)cos(A/6) = (1/2) * sin(A/3)

Answer: (1/2) * sin(A/3)

5. 20sin^2(4A)cos^2(4A):

Recall the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)

Using the double angle formula for sine: 20sin^2(4A)cos^2(4A) = 5 * sin^2(8A)

Answer: 5 * sin^2(8A)

6. sin(3A)cos(3A):

Recall the double angle formula for sine: sin(2θ) = 2sin(θ)cos(θ)

Using the double angle formula for sine: sin(3A)cos(3A) = (1/2) * sin(6A)

Answer: (1/2) * sin(6A)