Can someone please explan how I got this wrong?

Mathematics

1 answer:

Degger [83]3 years ago

6 0

$\bf \textit{Law of Cosines}\\\\
c^2 = a^2+b^2-(2ab)cos(C)\implies 
c = \sqrt{a^2+b^2-(2ab)cos(C)}
\\\\\\
\cfrac{a^2+b^2-c^2}{2ab}=cos(C)\implies cos^{-1}\left(\cfrac{a^2+b^2-c^2}{2ab}\right)=\measuredangle C\\\\
-------------------------------\\\\
\begin{cases}
x=24\\
a=25\\
b=19
\end{cases}\implies cos^{-1}\left(\cfrac{25^2+19^2-24^2}{2(25)(19)}\right)=\measuredangle x
\\\\\\
cos^{-1}\left( \cfrac{41}{95} \right)=\measuredangle x\implies 64.432194^o\approx \measuredangle x$

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timurjin [86]

Answer:

Proved

Step-by-step explanation:

$cos^4\alpha +sin^4\alpha =\frac{1}{4}(3+cos4\alpha )\\\\$

Take the Left Hand Side:

$cos^4\alpha +sin^4\alpha\\\\(cos^2\alpha )^2+(sin^2\alpha )^2\\\\(\frac{1+cos2\alpha }{2})^2+(\frac{1-cos2\alpha }{2})^2 \\\\\frac{1+2cos2\alpha +cos^22\alpha }{4}+\frac{1-2cos2\alpha +cos^22\alpha }{4} \\\\$

$\frac{1+2cos2\alpha +cos^22\alpha +1-2cos2\alpha +cos^22\alpha }{4} \\\\\frac{2+2cos^22\alpha }{4} \\\\\frac{1+cos^22\alpha }{2} \\\\\frac{1}{2}(1+cos^22\alpha )\\\\$