Question

cos^2(x)−sin((π)/(6)−x)sin((π)/(6)+x)

Answer 1

$\huge{\underline{\underline{\purple{♡Solution→}}}}$

${cos}^{2} x - \sin( \frac{\pi}{6} - x ) \sin( \frac{\pi}{6} + 1)$

We know that -

• π = 180

$= {cos}^{2} x - \sin( \frac{180}{6} - x) \sin( \frac{180}{6} + 1 ) \\ \\ = {cos}^{2} x - \sin(30 - x) \sin(30 + x)$

Using the formulas -

• $\sin(a - b) = \sin \: a \: cos \:b - cos \: a \: sin \: b$
• $\sin(a + b) = sin \: a \: cos \: b + cos \: as \: sin \: b$

$= {cos}^{2} x - (sin \: 30 \times cos \: x - cos \: 30 \times sin \: x)(sin \: 30 \times cos \: x + cos \: 30 \times sin \: x)$

But ,

• $sin \: 30 = \frac{1}{2}$
• $cos \: 30 = \frac{ \sqrt{3} }{2}$

$= {cos}^{2} x - ( \frac{1}{2} cos \: x - \frac{ \sqrt{3} }{2} sin \: x)( \frac{1}{2} cos \: x + \frac{ \sqrt{3} }{2} sin \: x)$

Using the identity -

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

$= {cos}^{2} x - (( \frac{1}{2}cos \: x)^{2} - ( \frac{\sqrt{3}}{2}sin \: x)^{2}))$

$={cos}^{2} x - ( \frac{1}{4}cos^{2} \: x - \frac{3}{4}sin^{2} \: x)$

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

$={cos}{^2}x-\frac{cos^{2}x-3sin^{2}x}{4}$

$=\frac{4cos^{2}x-cos^{2}x+3sin^{2}x}{4}$

$=\frac{3cos^{2}x+3sin^{2}x}{4}$

$=\frac{3(cos^{2}x+sin^{2}x)}{4}$

We know that -

• sin²x + cos²x = 1

$=\frac{3(1)}{4}$

$\boxed{ {cos}^{2} x - \sin( \frac{\pi}{6} - x ) \sin( \frac{\pi}{6} + 1)=\frac{3}{4}}$

$\rule{200}{1}$

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