3 years ago

Help please !!!!!!!!!!!

Mathematics

2 answers:

son4ous [18]3 years ago

6 0

It is definitely number one

blsea [12.9K]3 years ago

3 0

Inequality 1 I think

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Sin(60-theta)sin(60+theta)

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Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$

3 0

3 years ago

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