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
Find the value of (a) cos 120^(@) (b) sin 240^(@) (c) tan (-60^(
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6 𝑐𝑜𝑠60−sin30+sin^2 45+𝑐𝑜𝑠^2 45#shorts #sslcexam2022
Evaluate the following (i) `sin 60^ cos30^ + sin 30^ cos 60^` (ii) `2tan^2 45^ + cos^2 30^ - sin
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
Sin 150 Degrees - Find Value of Sin 150 Degrees
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Unit 2: Use reduction formulae to simplify trig ratios – National Curriculum (Vocational) Mathematics Level 3
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