Maths / Introduction to Trigonometry / Trigonometric Identities

QUESTION

If  $\dpi{80} \fn_phv cos\: \: \Theta -sin\: \: \Theta =\sqrt {2}\: \: sin\: \: \Theta ,$ then  $\dpi{80} \fn_phv cos \: \: \Theta +sin\: \: \Theta =$

 OPTIONS A. 0 B. $\dpi{80} \fn_phv \pm \sqrt {2}\: cos\: \: \Theta$ C. $\dpi{80} \fn_phv \pm \: \:\sqrt {2}\: \: sin\: \: \Theta$ D. 1
Right Option : B

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large cos\: \: \Theta -sin\: \: \Theta =\sqrt {2}\: \: sin\: \: \Theta ,$

Squaring Both Sides

$\dpi{120} \fn_jvn \large cos^2\Theta +sin^2\Theta- 2\;cos\;\Theta\;sin\Theta = 2\;sin^2\;\Theta$

$\dpi{120} \fn_jvn \large 1- 2\;cos\;\Theta\;sin\Theta = 2\;sin^2\;\Theta$

$\dpi{120} \fn_jvn \large - 2\;cos\;\Theta\;sin\Theta = 2\;sin^2\;\Theta -1$

$\dpi{120} \fn_jvn \large 2\;cos\;\Theta\;sin\Theta = 1-2\;sin^2\;\Theta$

$\dpi{120} \fn_jvn \large (cos\;\Theta+sin\;\Theta)^2=cos^2\Theta +sin^2\Theta+2\;cos\;\Theta\;sin\Theta$

$\dpi{120} \fn_jvn \large (cos\;\Theta+sin\;\Theta)^2=1+1-2\;sin^2\Theta$

$\dpi{120} \fn_jvn \large (cos\;\Theta+sin\;\Theta)^2=2-2\;sin^2\Theta$

$\dpi{120} \fn_jvn \large (cos\;\Theta+sin\;\Theta)^2=2(1-\;sin^2\Theta)$

$\dpi{120} \fn_jvn \large (cos\;\Theta+sin\;\Theta)^2=2(cos^2\Theta)$

$\dpi{120} \fn_jvn \large cos\;\Theta+sin\;\Theta=\pm \sqrt2\;cos\Theta$

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