Sin Π. limit_0^(π/2)(xsinxcosx)/(sin^4x+cos^4x)\ dxEvaluate∫(xsinxcosx)/(sin4x+cos4x)dx for x→[0π/2]#definiteintegrals #integration #samakalan #integral #rpkganit.

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Answer (1 of 7) You should notice that sin \pi lies in the II^{nd} quadrant hence the value is positive Using the property sin(A+B) = sinAcosB + cosAsinB \tag{1} Setting A=B we get sin(2A) = sinAcosA + cosAsinA \tag{2} = 2sinAcosA \tag{3} Setting A = \frac{\pi}{2} we have sin \pi = 2.

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The value of sin of 2pi is 0 ie sin 2π = 0 From trigonometric table we know the trigonometric ratios of standard angles 0 π/6 π/4 π/3 and π/2So this table doesn’t give us the value of sin of 2pi Usually to find the value of any trigonometric ratio of a nonstandard angle we use the reference angles and the quadrant in which the angle lies in.

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= (sin π/18) 2 + (sin π/9) 2 + (sin 7π/18) 2 + (sin 4π/9) 2 = sin 2 10 + sin 2 20 + sin 2 70 + sin 2 80 = [cos(90 10)] 2 + [cos(90 20)] 2 + sin 2 70 + sin 2 80 = cos 2 80 + cos 2 70 + sin 2 70 + sin 2 80 = sin 2 80 + cos 2 80 + sin 2 70 + cos 2 70 = 1 + 1 = 2 Apart from the stuff given above if you need any other stuff in math please use our google custom search here Kindly mail.