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rarely$66793$ - traduzione in olandese

rarely

adv. zeldzaam

trigonometric equation

$Diagram illustrating sum-to-product identities for sine and cosine. The blue right-angled triangle has angle <math>\theta</math> and the red right-angled triangle has angle <math>\varphi</math>. Both have a hypotenuse of length 1. Auxiliary angles, here called <math>p</math> and <math>q</math>, are constructed such that <math>p=(\theta+\varphi)/2</math> and <math>q=(\theta-\varphi)/2</math>. Therefore, <math>\theta = p+q </math> and <math>\varphi = p-q </math>. This allows the two congruent purple-outline triangles <math>AFG</math> and <math>FCE</math> to be constructed, each with hypotenuse <math>\cos q</math> and angle <math>p</math> at their base. The sum of the heights of the red and blue triangles is <math>\sin \theta + \sin \varphi</math>, and this is equal to twice the height of one purple triangle, i.e. <math>2 \sin p \cos q</math>. Writing <math>p</math> and <math>q</math> in that equation in terms of <math>\theta</math> and <math>\varphi</math> yields a sum-to-product identity for sine: <math>\sin \theta + \sin \varphi = 2 \sin\left( \frac{\theta + \varphi}{2} \right) \cos\left( \frac{\theta - \varphi}{2} \right)</math>. Similarly, the sum of the widths of the red and blue triangles yields the corresponding identity for cosine.$
$Sine power-reduction formula: an illustrative diagram. The shaded blue and green triangles, and the red-outlined triangle <math>EBD</math> are all right-angled and similar, and all contain the angle <math>\theta</math>. The hypotenuse <math>\overline{BD}</math> of the red-outlined triangle has length <math>2 \sin \theta</math>, so its side <math>\overline{DE}</math> has length <math>2 \sin^2 \theta</math>. The line segment <math>\overline{AE}</math> has length <math>\cos 2 \theta</math> and sum of the lengths of <math>\overline{AE}</math> and <math>\overline{DE}</math> equals the length of <math>\overline{AD}</math>, which is 1. Therefore, <math>\cos 2 \theta + 2 \sin^2 \theta = 1 </math>. Subtracting <math>\cos 2 \theta</math> from both sides and dividing by 2 by two yields the power-reduction formula for sine: <math> \sin^2 \theta = </math> ½<math> (1 - \cos (2\theta))</math>. The half-angle formula for sine can be obtained by replacing <math>\theta</math> with <math>\theta/2</math> and taking the square-root of both sides: <math>\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}.</math> Note that this figure also illustrates, in the vertical line segment <math>\overline{EB}</math>, that <math>\sin 2 \theta = 2 \sin \theta \cos \theta</math>.$
$Cosine power-reduction formula: an illustrative diagram. The red, orange and blue triangles are all similar, and the red and orange triangles are congruent. The hypotenuse <math>\overline{AD}</math> of the blue triangle has length <math>2 \cos \theta</math>. The angle <math>\angle DAE</math> is <math>\theta</math>, so the base <math>\overline{AE}</math> of that triangle has length <math>2 \cos^2 \theta</math>. That length is also equal to the summed lengths of <math>\overline{BD}</math> and <math>\overline{AF}</math>, i.e. <math>1 + \cos (2\theta)</math>. Therefore, <math>2 \cos^2\theta = 1 + \cos (2\theta)</math>. Dividing both sides by <math>2</math> yields the power-reduction formula for cosine: <math>\cos^2\theta =</math> ½<math>(1 + \cos (2\theta)) </math>. The half-angle formula for cosine can be obtained by replacing <math>\theta</math> with <math>\theta/2</math> and taking the square-root of both sides: <math>\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}.</math>$
$Diagram showing the angle difference identities for <math>\sin(\alpha - \beta)</math> and <math>\cos(\alpha - \beta)</math>.$
$Trigonometric functions and their reciprocals on the unit circle. All of the right-angled triangles are similar, i.e. the ratios between their corresponding sides are the same. For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them. The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. The triangle shaded blue illustrates the identity <math>1 + \cot^2\theta = \csc^2\theta</math>, and the red triangle shows that <math>\tan^2\theta + 1 = \sec^2\theta</math>.$
$Transformation of coordinates (''a'',''b'') when shifting the angle <math>\theta</math> in increments of <math>\frac{\pi}{2}</math>.$
$Transformation of coordinates (''a'',''b'') when shifting the reflection angle <math>\alpha</math> in increments of <math>\frac{\pi}{4}</math>.$
$2}} <math> \sin 2\theta</math>. Rotating the triangle does not change its area, so these two expressions are equal. Therefore, <math>\sin 2\theta = 2 \sin \theta \cos \theta</math>.$

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trigonometrische vergelijking (formule betreffende verhoudingen tussen zijden en driehoeken)