Question

高一数学三角比的求值题 sin^2 120°+cos90°+tan135°-cos^2(-330°)+sin(-240°

Answer 1

原式=sin²(180°-60°) + 0 + tan(180°-45°) - cos²[-(360°-30°)] - sin240°

=sin²60° + 0 - tan45° - cos²(-30°) - sin(270°-30°)

=(√3/2)² + 0 - 1 - (√3/2)² + cos30°

=3/4 - 1 - 3/4 + √3/2

=√3/2 - 1