Processing Math: Done
Solution 4.3:3d
From Förberedande kurs i matematik 1
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- | { | + | 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. |
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- | {{ | + | {| align="center" |
+ | | align="center" |[[Image:4_3_3_d-1.gif]] | ||
+ | | width="20px"| | ||
+ | | align="center" |[[Image:4_3_3_d-2.gif]] | ||
+ | |- | ||
+ | | align="center" |<small>Angle ''v''</small> | ||
+ | || | ||
+ | | align="center" |<small>Angle π/2 - ''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>}} |
Current revision
The expression for the angle 2−v
2
2−v
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Angle v | Angle π/2 - v |
Therefore, the angle 2−v
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and from exercise c, we know that 1−a2
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