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Solution for tan x is equal to cos x?

Tan(x) = Sin(x) / Cos(x) Hence Sin(x) / Cos(x) = Cos(x) Sin(x) = Cos^(2)[x] Sin(x) = 1 - Sin^(2)[x] Sin^(2)[x] + Sin(x) - 1 = 0 It is now in Quadratic form to solve for Sin(x) Sin(x) = { -1 +/-sqrt[1^(2) - 4(1)(-1)]} / 2(1) Sin(x) = { -1 +/-sqrt[5[} / 2 Sin(x) = {-1 +/-2.236067978... ] / 2 Sin(x) = -3.236067978...] / 2 Sin(x) = -1.61803.... ( This is unresolved as Sine values can only range from '1' to '-1') & Sin(x) = 1.236067978... / 2 Sin(x) = 0.618033989... x = Sin^(-1) [ 0.618033989...] x = 38.17270765.... degrees.


Is sin 2x equals 2 sin x cos x an identity?

YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx


Which trig expression is equal to sin (72 and Acirc and deg - a)?

The expression ( \sin(72^\circ - a) ) can be rewritten using the sine difference identity: [ \sin(72^\circ - a) = \sin(72^\circ) \cos(a) - \cos(72^\circ) \sin(a). ] Thus, ( \sin(72^\circ - a) ) is equal to ( \sin(72^\circ) \cos(a) - \cos(72^\circ) \sin(a) ).


What are five trigonometric identities?

All others can be derived from these and a little calculus: sin2x+cos2x=1 sec2x-tan2x=1 sin(a+b)=sin(a)cos(b)+sin(b)sin(a) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) eix=cos(x)+i*sin(x)


Sin3alfa equals to?

The expression (\sin(3\alpha)) can be expanded using the triple angle formula for sine, which is (\sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha)). This formula allows you to express (\sin(3\alpha)) in terms of (\sin(\alpha)).