Solution 1.1:2f
From Förberedande kurs i matematik 2
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| - | We can rewrite the function using a trigonometric addition formula | + | We can rewrite the function using a trigonometric addition formula, |
| + | {{Displayed math||<math>f(x) = \cos\Bigl(x+\frac{\pi}{3}\Bigr) = \cos x\cdot\cos \frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3}\,\textrm{.}</math>}} | ||
| - | + | If we now differentiate this expression, <math>\cos (\pi/3)</math> and <math>\sin (\pi/3)</math> are constants and we obtain | |
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| - | If we now differentiate this expression, | + | |
| - | <math>\cos | + | |
| - | and | + | |
| - | <math>\sin | + | |
| - | are constants and we obtain | + | |
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| + | {{Displayed math||<math>\begin{align} | ||
| + | f^{\,\prime}(x) | ||
| + | &= \frac{d}{dx}\,\Bigl(\cos x\cdot\cos\frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3} \Bigr)\\[5pt] | ||
| + | &= \cos\frac{\pi}{3}\cdot\frac{d}{dx}\,\cos x - \sin\frac{\pi}{3}\cdot\frac{d}{dx}\,\sin x\\[5pt] | ||
| + | &= \cos\frac{\pi}{3}\cdot (-\sin x) - \sin\frac{\pi}{3}\cdot\cos x\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
If we then use the addition formula in reverse, this gives | If we then use the addition formula in reverse, this gives | ||
| + | {{Displayed math||<math>\begin{align} | ||
| + | f^{\,\prime}(x) | ||
| + | &= -\Bigl(\sin x\cdot\cos\frac{\pi}{3} + \cos x\cdot\sin\frac{\pi}{3}\Bigr)\\[5pt] | ||
| + | &= -\sin\Bigl(x+\frac{\pi}{3}\Bigr)\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| - | <math>\begin{align} | ||
| - | & {f}'\left( x \right)=-\left( \sin x\centerdot \cos \frac{\pi }{3}+\cos x\centerdot \sin \frac{\pi }{3} \right) \\ | ||
| - | & =-\sin \left( x+\frac{\pi }{3} \right) \\ | ||
| - | \end{align}</math> | ||
| - | + | Note: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way. | |
Current revision
We can rewrite the function using a trigonometric addition formula,
| \displaystyle f(x) = \cos\Bigl(x+\frac{\pi}{3}\Bigr) = \cos x\cdot\cos \frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3}\,\textrm{.} |
If we now differentiate this expression, \displaystyle \cos (\pi/3) and \displaystyle \sin (\pi/3) are constants and we obtain
| \displaystyle \begin{align}
f^{\,\prime}(x) &= \frac{d}{dx}\,\Bigl(\cos x\cdot\cos\frac{\pi}{3} - \sin x\cdot\sin\frac{\pi}{3} \Bigr)\\[5pt] &= \cos\frac{\pi}{3}\cdot\frac{d}{dx}\,\cos x - \sin\frac{\pi}{3}\cdot\frac{d}{dx}\,\sin x\\[5pt] &= \cos\frac{\pi}{3}\cdot (-\sin x) - \sin\frac{\pi}{3}\cdot\cos x\,\textrm{.} \end{align} |
If we then use the addition formula in reverse, this gives
| \displaystyle \begin{align}
f^{\,\prime}(x) &= -\Bigl(\sin x\cdot\cos\frac{\pi}{3} + \cos x\cdot\sin\frac{\pi}{3}\Bigr)\\[5pt] &= -\sin\Bigl(x+\frac{\pi}{3}\Bigr)\,\textrm{.} \end{align} |
Note: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way.
