1.$\mathrm{sin}x{\mathrm{cos}}^{3}x-\mathrm{cos}x{\mathrm{sin}}^{3}x=1/4\mathrm{sin}4x$

2.$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\mathrm{cot}4x$

Darian Hubbard
2022-08-01
Answered

Solve.

1.$\mathrm{sin}x{\mathrm{cos}}^{3}x-\mathrm{cos}x{\mathrm{sin}}^{3}x=1/4\mathrm{sin}4x$

2.$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\mathrm{cot}4x$

1.$\mathrm{sin}x{\mathrm{cos}}^{3}x-\mathrm{cos}x{\mathrm{sin}}^{3}x=1/4\mathrm{sin}4x$

2.$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\mathrm{cot}4x$

You can still ask an expert for help

Monastro3n

Answered 2022-08-02
Author has **15** answers

1) Take out common factors:

$\mathrm{sin}x\mathrm{cos}x[{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x]=(1/4)\mathrm{sin}(4x)$

double angle identity:

$\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

So,

$\frac{\mathrm{sin}(2x)}{2}[\mathrm{cos}(2x)]=\frac{\mathrm{sin}(4x)}{4}$

$\frac{\mathrm{sin}(2x)\mathrm{cos}(2x)}{2}=\frac{\mathrm{sin}(4x)}{4}$

We already know that:

$\mathrm{sin}(2u)=\mathrm{sin}(u)\mathrm{cos}(u)$

if we let u = 2x:

$\mathrm{sin}[2(2x)]=\mathrm{sin}(2x)\mathrm{cos}(2x)$

$\mathrm{sin}(4x)=\mathrm{sin}(2x)\mathrm{cos}(2x)$

$\frac{\mathrm{sin}(4x)}{4}=\frac{\mathrm{sin}(4x)}{4}$

2) we know that $\mathrm{cos}x+\mathrm{cos}7x=2\mathrm{cos}(7x+x)/2\ast cos(7x-x)/2=2\mathrm{cos}4x\mathrm{cos}3x$

and $\mathrm{sin}x+\mathrm{sin}7x=2\mathrm{sin}(7x+x)/2\ast \mathrm{cos}(7x-x)/2=2\mathrm{sin}4x\mathrm{cos}3x$

$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\frac{2\mathrm{cos}4x\ast \mathrm{cos}3x+\mathrm{cos}4x}{2\mathrm{sin}4x\ast \mathrm{cos}3x+\mathrm{sin}4x}=\frac{\mathrm{cos}4x(2\mathrm{cos}3x+1)}{\mathrm{sin}4x(2\mathrm{cos}3x+1)}=\frac{\mathrm{cos}4x}{\mathrm{sin}4x}=\mathrm{cot}x$

$\mathrm{sin}x\mathrm{cos}x[{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x]=(1/4)\mathrm{sin}(4x)$

double angle identity:

$\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

So,

$\frac{\mathrm{sin}(2x)}{2}[\mathrm{cos}(2x)]=\frac{\mathrm{sin}(4x)}{4}$

$\frac{\mathrm{sin}(2x)\mathrm{cos}(2x)}{2}=\frac{\mathrm{sin}(4x)}{4}$

We already know that:

$\mathrm{sin}(2u)=\mathrm{sin}(u)\mathrm{cos}(u)$

if we let u = 2x:

$\mathrm{sin}[2(2x)]=\mathrm{sin}(2x)\mathrm{cos}(2x)$

$\mathrm{sin}(4x)=\mathrm{sin}(2x)\mathrm{cos}(2x)$

$\frac{\mathrm{sin}(4x)}{4}=\frac{\mathrm{sin}(4x)}{4}$

2) we know that $\mathrm{cos}x+\mathrm{cos}7x=2\mathrm{cos}(7x+x)/2\ast cos(7x-x)/2=2\mathrm{cos}4x\mathrm{cos}3x$

and $\mathrm{sin}x+\mathrm{sin}7x=2\mathrm{sin}(7x+x)/2\ast \mathrm{cos}(7x-x)/2=2\mathrm{sin}4x\mathrm{cos}3x$

$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\frac{2\mathrm{cos}4x\ast \mathrm{cos}3x+\mathrm{cos}4x}{2\mathrm{sin}4x\ast \mathrm{cos}3x+\mathrm{sin}4x}=\frac{\mathrm{cos}4x(2\mathrm{cos}3x+1)}{\mathrm{sin}4x(2\mathrm{cos}3x+1)}=\frac{\mathrm{cos}4x}{\mathrm{sin}4x}=\mathrm{cot}x$

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*a,b,c are positive I tried to base it on the solutions of a similar equation with the last term a 1 instead of a 2.