Question

Using the fact that $sin(A + B) = sinAcosB + cosAsinB$ and the differentiation, obtain the sum formula for cosines.

Answer 1

We solve this using product formula and chain rule of differentiation. Chain Rule of Differentiation: $\dfrac{{dy}}{{dx}}(f(g(x)) = f'(g(x)) \cdot g'(x) \cdot \dfrac{{d(x)}}{{dx}} = f'(g(x)) \cdot g'(x) \cdot 1$

Complete step-by-step answer:

Given that, $sin(A + B) = sinAcosB + cosAsinB$

Chain Rule of Differentiation: $\dfrac{{dy}}{{dx}}(f(g(x)) = f'(g(x)) \cdot g'(x) \cdot \dfrac{{d(x)}}{{dx}} = f'(g(x)) \cdot g'(x) \cdot 1$

Product formula of differentiation: $\dfrac{d}{{dx}}(fg) = f\dfrac{d}{{dx}}(g) + g\dfrac{d}{{dx}}(f)$

Differentiation of $sin(x)$ : $cos(x)$

Differentiation of $cos(x)$ : $-sin(x)$

Let A and B are the functions of $x$

Differentiate both sides of the above equation by using product rule on the right side and chain rule.

$\dfrac{d}{{dx}}(sin(A + B)) = \dfrac{d}{{dx}}(sinAcosB + cosAsinB)$

$\Rightarrow cos(A + B)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}) = [\dfrac{d}{{dx}}(sinAcosB)] + [\dfrac{d}{{dx}}(cosAsinB)]$

$\Rightarrow cos(A + B)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}) = [(sinA)\dfrac{d}{{dx}}(cosB) + (cosB)\dfrac{d}{{dx}}(sinA)] + [(cosA)\dfrac{d}{{dx}}(sinB) + (sinB)\dfrac{d}{{dx}}(cosA)]$

$\Rightarrow cos(A + B)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}) = sinA \cdot ( - sinB)\dfrac{{dB}}{{dx}} + cosB \cdot (cosA)\dfrac{{dA}}{{dx}} + cosA \cdot (cosB)\dfrac{{dB}}{{dx}} + sinB \cdot ( - sinA)\dfrac{{dA}}{{dx}}$

From Like terms take common factors:

$\Rightarrow cos(A + B)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}) = ( - sinA \cdot sinB)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}) + (cosA \cdot cosB)(\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}))$

Taking common $\dfrac{{dA}}{{dx}} + \dfrac{{dB}}{{dx}}$ from RHS and canceling, we get

$\Rightarrow cos(A + B) = cosA \cdot cosB - sinA \cdot sinB$

Hence, Above Equation is the required cosine formula.

Additional Note- Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable.

It is used in different ways, for example-

Suppose there is a moving car which travels from point A to B such that its position is changing with time and we can define its distance with a function $y=f(t)$ Therefore if we derive this function we will get the velocity of the car at time t.

The slope of a line, also called the gradient of the line, is a measure of its inclination. A line that is horizontal has slope 0, a line from the bottom left to the top right has a positive slope and a line from the top left to the bottom right has a negative slope.

Note: While solving make sure you take A and B functions of $x$ not constant because if you take them as a constant by differentiating you will get 0 on both sides because the derivative of constant function is 0. Also while solving do not forget to write $\dfrac{{dA}}{{dx}}$ and$\dfrac{{dB}}{{dx}}$. If you didn’t write this your question is wrong because of chain rule you should write this with them.