Jamison Rios

2022-07-02

Prove trigonometric identity and how to have ideas

This is the identity:

${\mathrm{cos}}^{2}(a)-{\mathrm{cos}}^{2}(b)=\frac{{\mathrm{tan}}^{2}(b)-{\mathrm{tan}}^{2}(a)}{(1+{\mathrm{tan}}^{2}(a))\cdot (1+{\mathrm{tan}}^{2}(b))}$

This is the identity:

${\mathrm{cos}}^{2}(a)-{\mathrm{cos}}^{2}(b)=\frac{{\mathrm{tan}}^{2}(b)-{\mathrm{tan}}^{2}(a)}{(1+{\mathrm{tan}}^{2}(a))\cdot (1+{\mathrm{tan}}^{2}(b))}$

earendil666h1

Beginner2022-07-03Added 10 answers

Starting from the Pythagorean Theorem:

$co{s}^{2}x+si{n}^{2}x=1$

Then dividing this by $co{s}^{2}x$ we have:

$1+ta{n}^{2}x=se{c}^{2}x$

Now let's look at the LHS:

$co{s}^{2}a-co{s}^{2}b=co{s}^{2}aco{s}^{2}b(\frac{1}{co{s}^{2}b}-\frac{1}{co{s}^{2}a})$

$=\frac{1}{se{c}^{2}ase{c}^{2}b}(se{c}^{2}b-se{c}^{2}a)$

From earlier we know how to substitute $se{c}^{2}x$ and have

$=\frac{1}{(1+ta{n}^{2}a)(1+ta{n}^{2}b)}(1+ta{n}^{2}b)-(1+ta{n}^{2}a)$

Or simplified:

$\frac{ta{n}^{2}b-ta{n}^{2}a}{(1+ta{n}^{2}a)(1+ta{n}^{2}b)}$

The identities often revert back to P.T. at some point and is often a good place to start.

$co{s}^{2}x+si{n}^{2}x=1$

Then dividing this by $co{s}^{2}x$ we have:

$1+ta{n}^{2}x=se{c}^{2}x$

Now let's look at the LHS:

$co{s}^{2}a-co{s}^{2}b=co{s}^{2}aco{s}^{2}b(\frac{1}{co{s}^{2}b}-\frac{1}{co{s}^{2}a})$

$=\frac{1}{se{c}^{2}ase{c}^{2}b}(se{c}^{2}b-se{c}^{2}a)$

From earlier we know how to substitute $se{c}^{2}x$ and have

$=\frac{1}{(1+ta{n}^{2}a)(1+ta{n}^{2}b)}(1+ta{n}^{2}b)-(1+ta{n}^{2}a)$

Or simplified:

$\frac{ta{n}^{2}b-ta{n}^{2}a}{(1+ta{n}^{2}a)(1+ta{n}^{2}b)}$

The identities often revert back to P.T. at some point and is often a good place to start.

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