Solution 4.2:5d
From Förberedande kurs i matematik 1
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- | {{ | + | By subtracting 360° from 495°, we do not change the value of the tangent, |
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- | {{ | + | {{Displayed math||<math>\tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.}</math>}} |
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+ | We know from exercise a that <math>\cos 135^{\circ} = -1/\!\sqrt{2}</math> and <math>\sin 135^{\circ} = 1/\!\sqrt{2}\,</math>, which gives | ||
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+ | {{Displayed math||<math>\tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}} |
Current revision
By subtracting 360° from 495°, we do not change the value of the tangent,
\displaystyle \tan 495^{\circ} = \tan (495^{\circ} - 360^{\circ}) = \tan 135^{\circ}\,\textrm{.} |
We know from exercise a that \displaystyle \cos 135^{\circ} = -1/\!\sqrt{2} and \displaystyle \sin 135^{\circ} = 1/\!\sqrt{2}\,, which gives
\displaystyle \tan 135^{\circ} = \frac{\sin 135^{\circ}}{\cos 135^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.} |