6. sin150 Sin(90+60)=+cos60∘=+1/2 sin(180−30)=+sin30∘+1/2
7cos120∘cos(90+30)=−sin30∘=+1/2cos(180∘−60)=−cos60∘=−1/2
cos(90+60)=−sin60∘=−8/2cos(180∘−30)=−cos30=−3/2
cos150
Video solution 1: 6. sin150 Sin(90+60)=+cos60∘=+1/2 sin(180−30)=+sin30∘+1/2
7cos120∘cos(90+30)=−sin30∘=+1/2cos(180∘−60)=−cos60∘=−1/2
cos(90+60)=−sin60∘=−8/2cos(180∘−30)=−cos30=−3/2
cos150
Video solution 2: 6. sin150 Sin(90+60)=+cos60∘=+1/2 sin(180−30)=+sin30∘+1/2
7cos120∘cos(90+30)=−sin30∘=+1/2cos(180∘−60)=−cos60∘=−1/2
cos(90+60)=−sin60∘=−8/2cos(180∘−30)=−cos30=−3/2
cos150
Show that: 1-sin60/ cos60 = tan60 -1/tan60 +1 - Sarthaks eConnect
SOLVED: 21. sin 60^∘(csc 60^∘-sin 60^∘) 22. (1+cot ^2 30^∘)/(1+tan ^2 30^∘) 23. sin 60^∘cos 60^∘tan 60^∘cot 60^∘sec 60^∘csc 60^∘ 24. sin 30^∘cos 60^∘+cos 30^∘sin 60^∘ 25. sec 60^∘-sin 30^∘+cot 45^∘ 26.
What is the value of sin150? - Quora
SOLVED: -cos 30^∘=(√(3))/(2) -cos 60^∘=(1)/(2) -cos 90^∘=0 -cos 120^∘=(180^∘ -60^∘)=-cos 60^∘=-(1)/(2) -cos 180^∘=cos(180^∘-80^∘)=-cos 30^∘=(-√(3))/(2) -cos 180^∘=cos(180^∘-0^∘)=-cos 60^∘=-(1)/(2) -cos 210^∘=cos(180^∘+30^∘)=-30
Why sin30=cos60=1/2, sin60=cos30=square root 3 over 2, tan30=1/square root 3, tan60=square root 3 ?
SOLVED: -cos 30^∘=(√(3))/(2) -cos 60^∘=(1)/(2) -cos 90^∘=0 -cos 120^∘=(180^∘ -60^∘)=-cos 60^∘=-(1)/(2) -cos 180^∘=cos(180^∘-80^∘)=-cos 30^∘=(-√(3))/(2) -cos 180^∘=cos(180^∘-0^∘)=-cos 60^∘=-(1)/(2) -cos 210^∘=cos(180^∘+30^∘)=-30
If sin 600° cos 30° + cos 120° sin 150° = k, then k = (a) 0 (b) 1 (c) −1 (d) None of these
Ex 8.2, 1 Class 10 - Evaluate (i) sin 60° cos 30° + sin 30° cos 60°
evaluate: Sin 60°/cos^2 45°- cot30°+ 15 cos 90°
