![\bf cos(\alpha+\beta)cos(\alpha-\beta)\\\\ -----------------------------\\\\ \textit{using the sum identities}\\\\\ \begin{array}{ccccccll} [cos(\alpha)cos(\beta)&-&sin(\alpha)sin(\beta)]&[cos(\alpha)cos(\beta)&+&sin(\alpha)sin(\beta)]\\ a&-&b&a&+&b \end{array}\\\\ \textit{notice above, is just a difference of squares, thus} \\\\\\\ [cos(\alpha)cos(\beta)]^2-[sin(\alpha)sin(\beta)]^2\\\\ \textit{now let us distribute the exponents}\\\\](/tpl/images/0444/8045/be4b7.png)
![\bf cos^2(\alpha)cos^2(\beta)-sin^2(\alpha)sin^2(\beta)\\\\ -----------------------------\\\\ \textit{now, let us use the pythagorean identity of }\\\\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -----------------------------\\\\\ [ 1-sin^2(\alpha)][ 1-sin^2(\beta)]-sin^2(\alpha)sin^2(\beta) \\\\\\\ 1-sin^2(\beta)-sin^2(\alpha)\underline{+sin^2(\alpha)sin^2(\beta)-sin^2(\alpha)sin^2(\beta)} \\\\\\ \underline{1-sin^2(\beta)}-sin^2(\alpha)\implies cos^2(\beta)-sin^2(\alpha)](/tpl/images/0444/8045/b78ec.png)
Attached is a screenshot of another problem i am having issues with. me
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Mathematics, 06.01.2020 23:31, peno211
Attached is a screenshot of another problem i am having issues with. me
Answers: 2
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Me with this one question, and i'll upvote the brainliest answer
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