Difficult
Using the fact that $\sin (A+B)=\sin A \cos B+\cos A \sin B$ and the differentiation,obtain the sum formula for cosines.
Given the identity: $\sin (A+B)=\sin A \cos B+\cos A \sin B$
Differentiating both sides with respect to $x$,we apply the product rule and chain rule:
$\frac{d}{d x}[\sin (A+B)]=\frac{d}{d x}(\sin A \cos B)+\frac{d}{d x}(\cos A \sin B)$
$\cos (A+B) \cdot \frac{d}{d x}(A+B) = \cos B \cdot \frac{d}{d x}(\sin A) + \sin A \cdot \frac{d}{d x}(\cos B) + \sin B \cdot \frac{d}{d x}(\cos A) + \cos A \cdot \frac{d}{d x}(\sin B)$
Using the chain rule,$\frac{d}{dx}(\sin A) = \cos A \frac{dA}{dx}$,$\frac{d}{dx}(\cos A) = -\sin A \frac{dA}{dx}$,etc.:
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos B (\cos A \frac{dA}{dx}) + \sin A (-\sin B \frac{dB}{dx}) + \sin B (-\sin A \frac{dA}{dx}) + \cos A (\cos B \frac{dB}{dx})$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos A \cos B \frac{dA}{dx} - \sin A \sin B \frac{dB}{dx} - \sin A \sin B \frac{dA}{dx} + \cos A \cos B \frac{dB}{dx}$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos A \cos B \left( \frac{dA}{dx} + \frac{dB}{dx} \right) - \sin A \sin B \left( \frac{dA}{dx} + \frac{dB}{dx} \right)$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = (\cos A \cos B - \sin A \sin B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right)$
Canceling the common term $\left( \frac{dA}{dx} + \frac{dB}{dx} \right)$,we get:
$\cos (A+B) = \cos A \cos B - \sin A \sin B$
Differentiating both sides with respect to $x$,we apply the product rule and chain rule:
$\frac{d}{d x}[\sin (A+B)]=\frac{d}{d x}(\sin A \cos B)+\frac{d}{d x}(\cos A \sin B)$
$\cos (A+B) \cdot \frac{d}{d x}(A+B) = \cos B \cdot \frac{d}{d x}(\sin A) + \sin A \cdot \frac{d}{d x}(\cos B) + \sin B \cdot \frac{d}{d x}(\cos A) + \cos A \cdot \frac{d}{d x}(\sin B)$
Using the chain rule,$\frac{d}{dx}(\sin A) = \cos A \frac{dA}{dx}$,$\frac{d}{dx}(\cos A) = -\sin A \frac{dA}{dx}$,etc.:
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos B (\cos A \frac{dA}{dx}) + \sin A (-\sin B \frac{dB}{dx}) + \sin B (-\sin A \frac{dA}{dx}) + \cos A (\cos B \frac{dB}{dx})$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos A \cos B \frac{dA}{dx} - \sin A \sin B \frac{dB}{dx} - \sin A \sin B \frac{dA}{dx} + \cos A \cos B \frac{dB}{dx}$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = \cos A \cos B \left( \frac{dA}{dx} + \frac{dB}{dx} \right) - \sin A \sin B \left( \frac{dA}{dx} + \frac{dB}{dx} \right)$
$\cos (A+B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right) = (\cos A \cos B - \sin A \sin B) \left( \frac{dA}{dx} + \frac{dB}{dx} \right)$
Canceling the common term $\left( \frac{dA}{dx} + \frac{dB}{dx} \right)$,we get:
$\cos (A+B) = \cos A \cos B - \sin A \sin B$
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MediumIIT 1988
View SolutionGive the correct order of initials $T$ or $F$ for the following statements. Use $T$ if the statement is true and $F$ if it is false.
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Statement-$2$: Let $f: (a, b) \rightarrow R, c \in (a, b)$. Then $f$ cannot have both a local maximum and a point of inflection at $x = c$.
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Statement-$1$: If $f: R \rightarrow R$ and $c \in R$ is such that $f$ is increasing in $(c - \delta, c)$ and $f$ is decreasing in $(c, c + \delta)$,then $f$ has a local maximum at $c$. Where $\delta$ is a sufficiently small positive quantity.
Statement-$2$: Let $f: (a, b) \rightarrow R, c \in (a, b)$. Then $f$ cannot have both a local maximum and a point of inflection at $x = c$.
Statement-$3$: The function $f(x) = x^2 |x|$ is twice differentiable at $x = 0$.
Statement-$4$: Let $f: [c - 1, c + 1] \rightarrow [a, b]$ be a bijective map such that $f$ is differentiable at $c$ and $f'(c) \neq 0$,then $f^{-1}$ is also differentiable at $f(c)$.
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Let $D$ be the domain of a twice differentiable function $f$. For all $x \in D, f^{\prime \prime}(x)+f(x)=0$ and $f(x)=\int g(x) \, dx + \text{constant}$. If $h(x)={f(x)}^2+{g(x)}^2$ and $h(0)=5$,then $h(2015)-h(2014)$ is equal to
DifficultAP EAMCET 2015
View SolutionIf a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:
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