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

2.$\frac{1-\mathrm{cos}(X)}{1+\mathrm{cos}(X)}={\mathrm{tan}}^{2}(\frac{x}{2})$

Kade Reese
2022-07-26
Answered

Prove the given identities.

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

2.$\frac{1-\mathrm{cos}(X)}{1+\mathrm{cos}(X)}={\mathrm{tan}}^{2}(\frac{x}{2})$

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

2.$\frac{1-\mathrm{cos}(X)}{1+\mathrm{cos}(X)}={\mathrm{tan}}^{2}(\frac{x}{2})$

You can still ask an expert for help

Cheyanne Charles

Answered 2022-07-27
Author has **13** answers

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

we have $({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x)=1\text{}and\text{}{\mathrm{cos}}^{2}x=2{\mathrm{cos}}^{2}x-1$

$={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$

$={\mathrm{cos}}^{2}x-(1-{\mathrm{cos}}^{2}x)$

$=2{\mathrm{cos}}^{2}x-1$

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

2) We have the trignometric relations

$1+\mathrm{cos}(2t)=2{\mathrm{cos}}^{2}t$

$1-\mathrm{cos}(2t)=2{\mathrm{sin}}^{2}t$

Putting 2t = x, we get

$1+\mathrm{cos}(x)=2{\mathrm{cos}}^{2}(x/2)\to 1$

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

$2/1\Rightarrow {\mathrm{tan}}^{2}(x/2)$

we have $({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x)=1\text{}and\text{}{\mathrm{cos}}^{2}x=2{\mathrm{cos}}^{2}x-1$

$={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$

$={\mathrm{cos}}^{2}x-(1-{\mathrm{cos}}^{2}x)$

$=2{\mathrm{cos}}^{2}x-1$

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

2) We have the trignometric relations

$1+\mathrm{cos}(2t)=2{\mathrm{cos}}^{2}t$

$1-\mathrm{cos}(2t)=2{\mathrm{sin}}^{2}t$

Putting 2t = x, we get

$1+\mathrm{cos}(x)=2{\mathrm{cos}}^{2}(x/2)\to 1$

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

$2/1\Rightarrow {\mathrm{tan}}^{2}(x/2)$

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