What is the value of Sin345?

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Salman Ahmad Siddiqui

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

Sin(345)=sin(360+(-15))=sin(-15)=-sin(15) =-Sin(45-30) Sin(A-B)=sinA cos(B) - cos(A)sin(B) =-Sin(15)=-sin(45-30) =- =- =- =- =- =-0.2588
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Sin(345)=sin(360+(-15))=sin(-15)=-sin(15) =-Sin(45-30) Sin(A-B)=sinA cos(B) - cos(A)sin(B) =-Sin(15)=-sin(45-30) =-[Sin(45)cos(30) -cos(45)sin(30)] =-[(1/√2)(√3/2) - (1/√2)(1/2)] =-[(√3 -1)/(2√2)] =-[(1.732-1)/(2*1.414)] =-[(0.732)/2.828] =-0.2588 read less
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Teachar

Assuming we're in degrees, we have sin(345) = sin(300 + 45)....and using sin(A + B) = sinAcosB + sinBcosA, we have sin300cos45 + sin45cos300 = (-√3/2)(1/√2) + (1/√2)(1/2) = -√3/ (2√2) + 1/(2√2) = / (2√2) and rationalizing the denominator, we have √2...
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Assuming we're in degrees, we have sin(345) = sin(300 + 45)....and using sin(A + B) = sinAcosB + sinBcosA, we have sin300cos45 + sin45cos300 = (-√3/2)(1/√2) + (1/√2)(1/2) = -√3/ (2√2) + 1/(2√2) = [ 1 - √3 ] / (2√2) and rationalizing the denominator, we have √2 [ 1 - √3] / 4 = (√2 - √6) / 4 read less
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sin(300+45) = sin300cos45 + sin45cos300 (-√3/2)(1/√2) + (1/√2)(1/2) ...
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Sin(360-15)=- sin15=-0.25
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Sin(360-15)=-sin15=-(√3-1)/2√2
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