Question

2sin^2 (8×1/2)+4cos16sin (7×1/2)sin (8×1/2)+cos32 =?​

Answer 1

we know that sin2A-sin2B = sin (A+B)sin(A-B)π/8 + a/2+π/8 - a/2) sin(π/8 + a/2-π/8+ a/2)sin^2[π÷8+A÷2]-sin^2[π÷8-A÷2]=sin(π/4) sin(a)=sin(=(1/ ) sina .......

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Answer 2

Answer:

√3+1/2√2

Step-by-step explanation:

1-cos17°+2cos16°[2sin(7 1/2°)sin(8 1/2°)]+cos32°

{1-cos17°,since 2sin²A/2=1-cosA}

1-cos17°+2cos16°[cos(15/2°-17/2°)-cos(15/2°+17/2°)]+cos32°

{since cos(A-B)-cos(A+B)=2sinAsinB}

1-sin17°+2cos16°[cos1°-cos16°]+cos32°

1-cos17°+cos(16°+1°)+cos(16°-1°)-2cos²16°+cos32°

{cos(A+B)+cos(A-B)= 2cosAcosB}

1-cos17°+cos17°+cos15°-1-cos32°+cos32°

{since 2cos²A/2=1+cosA}

cos15°=√3+1/2√2