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2 years ago

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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Read 2 more answers

Directions : Factor each of the following Differences of two squares and write your answer together with solution

N76 [4]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Factor each of the following differences of two squares and write your answer together with solution.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf \: x ^ { 2 } - 36$

Rewrite $\sf\:x^{2}-36 as x^{2}-6^{2}$ . The difference of squares can be factored using the rule: $\sf\: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\boxed{ \boxed{ \bf\left(x-6\right)\left(x+6\right) }}$

__________________

$\sf \: 49 - x ^ { 2 }$

Rewrite 49-x² as 7²-x². The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\sf \: \left(7-x\right)\left(7+x\right)$

Reorder the terms.

$\boxed{ \boxed{ \bf\left(-x+7\right)\left(x+7\right) }}$

__________________

$\sf \: 81 - c ^ { 2 }$

Rewrite 81-c²as 9²-c². The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\sf\left(9-c\right)\left(9+c\right)$

Reorder the terms.

$\boxed{ \boxed{ \bf\left(-c+9\right)\left(c+9\right) }}$

__________________

$\sf \: m ^ { 2 } n ^ { 2 } - 1$

Rewrite m²n² - 1 as $\sf\left(mn\right)^{2}-1^{2}$ . The difference of squares can be factored using the rule: $\sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .