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vaieri [72.5K]
2 years ago
9

Establish the identity. (2 cos 0-6 sin 0)² + (6 cos 0+2 sin 0)2 = 40

Mathematics
1 answer:
kogti [31]2 years ago
8 0

Rewriting the left-hand side as follows,

(2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40

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Answer:

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Step-by-step explanation:

The given expression is

\sqrt[3]{135}

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Using this property we get

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\sqrt[3]{135}=(3^3)^{\frac{1}{3}}\times (5)^{\frac{1}{3}}      [\because (ab)^x=a^xb^x]

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\huge\text{Answer:}

\huge\mathrm{0.2x+0.3}\checkmark

\huge\text{Solution:}

Hi there, hope you are having a great day! :)

<em />\bigstar\star\star\sf{What\:is\:the\:Distributive\:Property?}

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Feel free to ask if you have any doubts.

\bf{-MistySparkles^**^*

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