From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	The expression for the angle <math>\pi/2 - v</math> differs from <math>\pi/2</math> by as much as <math>-v</math> differs from <math>0</math>. This means that <math>\pi/2-v</math> makes the same angle with the positive ''y''-axis as <math>-v</math> makes with the positive ''x''-axis.
-	<center> [[Image:4_3_3d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{| align="center"
-	[[Image:4_3_3_d.gif|center]]	+	| align="center" |[[Image:4_3_3_d-1.gif]]
		+	| width="20px"|&nbsp;
		+	| align="center" |[[Image:4_3_3_d-2.gif]]
		+	|-
		+	| align="center" |<small>Angle&nbsp;''v''</small>
		+	||
		+	| align="center" |<small>Angle&nbsp;π/2&nbsp;-&nbsp;''v''</small>
		+	|}
		+
		+	Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.
		+
		+	{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
		+
		+	and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
		+
		+	{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}

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Current revision

The expression for the angle \displaystyle \pi/2 - v differs from \displaystyle \pi/2 by as much as \displaystyle -v differs from \displaystyle 0. This means that \displaystyle \pi/2-v makes the same angle with the positive y-axis as \displaystyle -v makes with the positive x-axis.

Angle v		Angle π/2 - v

Therefore, the angle \displaystyle \pi/2 - v has a y-coordinate which is equal to the x-coordinate for the angle v, i.e.

\displaystyle \sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v

and from exercise c, we know that \displaystyle \cos v = \sqrt{1-a^2}\,,

\displaystyle \sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}