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Maths-

General

Easy

Question

$I f sin invisible function application 2 theta equals cos invisible function application 3 theta blank a n d blank theta$ is an acute angle, then $sin invisible function application theta$ equals

$\frac{\sqrt{5} - 1}{4}$
$- (\frac{\sqrt{5} - 1}{4})$
$\frac{\sqrt{5} + 1}{4}$
$\frac{- \sqrt{5} - 1}{4}$

The correct answer is: $fraction numerator square root of 5 minus 1 over denominator 4 end fraction$

$sin invisible function application 2 theta equals cos invisible function application 3 theta blank rightwards double arrow 2 sin invisible function application theta cos invisible function application theta equals 4 cos to the power of 3 end exponent invisible function application theta minus 3 cos invisible function application theta$
$rightwards double arrow 2 sin invisible function application theta equals 4 open parentheses negative sin to the power of 2 end exponent invisible function application theta close parentheses minus 3 rightwards double arrow 4 sin to the power of 2 end exponent invisible function application theta plus 2 sin invisible function application theta minus 1 equals 0$
$rightwards double arrow sin invisible function application theta equals fraction numerator square root of 5 minus 1 over denominator 4 end fraction$

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Physics-General

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