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What equals 1 - cos squared x divided by cos squared x?
tan^2(x) Proof: cos^2(x)+sin^2(x)=1 (Modified Pythagorean theorem) sin^2(x)=1-cos^2(x) (Property of subtraction) cos^2(x)-1/cos^2(x)=? sin^2(x)/cos^2(x)=? (Property of substitution) sin(x)/cos(x) * sin(x)/cos(x) = tan(x) * tan(x) (Definition of tanget) = tan^2(x)
Sine divided by cosine?
Trig identity... sin/cos = tangent
What is equivalent to sin 2 csc 2 - sin 2?
sin2csc2-sin2 (using the fact that the sin is the reciprocal of csc) = 1-sin2
What is sine squared?
Answer 1 Put simply, sine squared is sinX x sinX. However, sine is a function, so the real question must be 'what is sinx squared' or 'what is sin squared x': 'Sin(x) squared' would be sin(x^2), i.e. the 'x' is squared before performing the function sin. 'Sin squared x' would be sin^2(x) i.e. sin squared times sin squared: sin(x) x sin(x). This can also be written as (sinx)^2 but means exactly the same. Answer 2 Sine squared is sin^2(x). If the power was placed like this sin(x)^2, then the X is what is being squared. If it's sin^2(x) it's telling you they want sin(x) times sin(x).
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
