Mindblanked on these two questions any help appreciated (1 Viewer)

lmao1010 · Feb 5, 2022

Hey there if any1 cud just guide me on how to approach these questions haven't touched them since late yr11 so a bit rusty rn

5uckerberg · Feb 5, 2022

lmao1010 said:
Hey there if any1 cud just guide me on how to approach these questions haven't touched them since late yr11 so a bit rusty rn

For this question $-\tan{2x}\tan{x}=\frac{\cos{3x}-\cos{x}}{\cos{x}+\cos{3x}}$ , we can start working on this question knowing we need a $2x$ trig term and a $x$ trig term.

Now the mechanical step.

$\cos{3x}-\cos{x}=\cos{\left(2x+x\right)}-\cos{\left(2x-x\right)}$
$\cos{x}+\cos{3x}=\cos{\left(2x-x\right)}+\cos{\left(2x+x\right)}$

Noting that
$\cos{\left(2x+x\right)}-\cos{\left(2x-x\right)}=\cos{2x}\cos{x}-\sin{2x}\sin{x}-\left(\cos{2x}\cos{x}+\sin{2x}\sin{x}\right)$
$\cos{\left(2x-x\right)}+\cos{\left(2x+x\right)}=\cos{2x}\cos{x}+\sin{2x}\sin{x}+\left(\cos{2x}\cos{x}-\sin{2x}\sin{x}\right)$

Knowing this
$\frac{\cos{3x}-\cos{x}}{\cos{x}+\cos{3x}}=\frac{-2\sin{2x}\sin{x}}{2\cos{2x}\cos{x}}=-\tan{2x}\tan{x}$

5uckerberg · Feb 5, 2022

Have a look at the working out. If you need assistance someone will help interpret that information.

Mindblanked on these two questions any help appreciated (1 Viewer)

lmao1010

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5uckerberg

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