How to prove sin3θ/sinθ + cos3θ/cosθ = 4cos2θ? Past paper question...Help...?
回答 (1)
2015-01-31 5:00 pm
[sin(3*x)/sin x] + [cos(3*x)/cosx]
Common denominator...
[(sin (3*x) * cos x) + (cos (3*x) * sin x)]/(cos x*sin x)
Sum of sines in numerator, remember formula for sin 2*z in denominator....
sin(4*x)/[(1/2)*sin(2*x)]
2*sin(4*x)/sin(2*x)
Write 4*x = sin(2*2*x) and use sum of sines formula
2*(2*sin(2*x)*cos(2*x))/sin(2*x)
4*cos(2*x)
Common denominator...
[(sin (3*x) * cos x) + (cos (3*x) * sin x)]/(cos x*sin x)
Sum of sines in numerator, remember formula for sin 2*z in denominator....
sin(4*x)/[(1/2)*sin(2*x)]
2*sin(4*x)/sin(2*x)
Write 4*x = sin(2*2*x) and use sum of sines formula
2*(2*sin(2*x)*cos(2*x))/sin(2*x)
4*cos(2*x)
收錄日期: 2021-04-15 18:08:56
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