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for this one, do i make it into double angle formula first or do i do something else? (1 Viewer)
- Thread starter yashbb
- Start date
how would i do that
Or... by calculus:
^2} \\ &= \cfrac{2\sin^2{2x} - 2\cos{2x} + 2\cos^2{2x}}{\sin^2{2x}} \\ &= \cfrac{2(1 - \cos{2x})}{\sin^2{2x}} \qquad \qquad \text{as $\sin^2{2x} + \cos^2{2x} =1$} \\ &= \cfrac{2 \times 2\sin^2{x}}{(2\sin{x}\cos{x})^2} \qquad \qquad \text{since $\cos{2x} = 1 - 2\sin^2{x}$} \\ &= \cfrac{4\sin^2{x}}{4\sin^2{x}\cos^2{x}} \\ &= \cfrac{1}{\cos^2{x}} \\ &= \sec^2{x} \\ \text{Thus, LHS}\ &= \int sec^2{x}\ dx \\ &= \tan{x} + C \qquad \text{for some constant $C$} \\ \text{Now, testing $x = \cfrac{\pi}{4}$ :} \quad \cfrac{1-\cos{\frac{2\pi}{4}}}{\sin{\frac{2\pi}{4}}} &= \tan{\cfrac{\pi}{4}} + C \\ \cfrac{1-0}{1} &= 1 + C \\ C &= 0 \\ \text{Thus, LHS}\ &= \tan{x} \\ &=\ \text{RHS} \end{align*})
The LHS can also be re-written as
before differentiating.
Also, the non-calculus approach is definitely easier in this case!
(Especially if I went more carefully through the problems raised by discontinuities...)
The LHS can also be re-written as
Also, the non-calculus approach is definitely easier in this case!
(Especially if I went more carefully through the problems raised by discontinuities...)