tan(-150°)cos(-570°)cos(-1140°)/cot(-240°)sin（-690°)

tan(-150°)cos(-570°)cos(-1140°)/cot(-240°)sin（-690°)

tan(-150°)cos(-570°)cos(-1140°)/cot(-240°)sin（-690°)= tan(-150°)cos(150°)cos(-60°)/cot(120°)sin（30°)= -tan(30°)cos(30°)cos(60°)/-cot(60°)sin（30°)= sin(30°)cos(30°)cos(60°)sin(60°...

Calculate (1) sin 420 degrees * cos 750 degrees + sin (- 330 degrees) * cos (- 660 degrees) Tan 675 degrees + Tan 765 degrees - Tan (- 330 degrees) + Tan (- 690 degrees)
(3) Sin25tt / 6 + cos25tt / 3 + Tan (- 25tt / 4) process

Solution (1): original formula = sin (360 ° + 60 °) × cos (2 × 360 ° + 30 °) - Sin 330 ° × cos 660 °
=sin60°×cos30°-sin(360°-30°)×cos(3×360°-60°)
=sin60°×cos30°-(-sin30°)×cos(-60°)
=sin60°×cos30°+sin30°×cos60°
=√3/2×√3/2+1/2×1/2
=3/4+1/4
=1
Solution (2): original formula = Tan (2 × 360 ° - 45 °) + Tan (2 × 360 ° + 45 °) + Tan 330 ° - Tan 690 °
=-tan45°+tan45°+tan(360°-30°)-tan(2×360°-30°)
=-tan30°-(-tan30°)
=-tan30°+tan30°
=0
Solution 3: the original formula = sin [4 π + (π / 6)] + cos [8 π + (π / 3)] + Tan [- 6 π + (- π / 4)]
=sin(π/6)+cos(π/3)+tan(-π/4)
=sin(π/6)+cos(π/3)-tan(π/4)
=1/2+1/2-1
=0

sin420cos330+sin(-690)cos(-660)=

sin420cos330+sin(-690)cos(-660)=sin60cos30+sin30cos60=sin(30+60)=1

Is sin (a-b) = sin (a + b)?

sin(a-b)=sinacosb-sinbcosa
sin(a+b)=sinacosb+sinbcosa
So it's not equal