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$prove \: that \\ { \cos( \alpha ) }^{3} \cos(3 \alpha ) + { \sin( \alpha ) }^{3} \sin(3 \alpha ) = { \cos(2 \alpha ) }^{3}$








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Answer 1

Appropriate question :

$\bf \: prove \: that \: \\ \rm \: {cos}^{3} \alpha . \: cos3 \alpha + {sin}^{ 3} \alpha .sin \: 3 \alpha = {cos}^{3} \: 3 \alpha$

Solution :

${\boxed{\boxed{\begin{array}{cc} \rm \: L.H.S = {cos}^{3} \alpha \: . \: cos \: 3 \alpha + {sin}^{3} \alpha \: . \: sin \: 3 \alpha \\ \\ \rm = \frac{1}{4} . (4 \: {cos}^{3} \alpha ).cos \: 3 \alpha + \frac{1}{4} . (4 \: sin {}^{3} \alpha ).sin \: 3 \alpha \\ \\ \orange{{\boxed{\begin{array}{cc} \text{we \: know \: that : } \\ \\ \to \: \rm \: cos \: 3x = 4 {cos}^{3}x - 3 \: cos \: x \\ \therefore \rm \: 4 \: {cos}^{3} x = cos \: 3x + 3 \: cos \: x \\ \\ \to \rm \: sin \: 3x = 3 \: sin \: x - 4 {sin}^{3} x \\ \therefore \: \rm \: 4 {sin}^{3} x = 3 \: sin \: x - sin \: 3x\end{array}}}} \\ \pink{ \text{apply \: this}} \\ \\ \rm = \frac{1}{4}.(cos \: 3 \alpha + 3 \: cos \: \alpha ).cos \: 3 \alpha + \frac{1}{4} .(3 \: sin \: \alpha - sin \: 3 \alpha ) .sin \: 3 \alpha \\ \\ \rm = \frac{1}{4} .( {cos}^{2} \: 3\alpha + 3 \: cos \: \alpha .cos \: 3 \alpha ) + \frac{1}{4} .(3 \: sin \: \alpha \: sin \: 3 \alpha - {sin}^{2} \: 3 \alpha ) \\ \\ \rm = \frac{1}{4} .( {cos}^{2} \: 3\alpha + 3 \: cos \: \alpha .cos \: 3 \alpha + 3 \: sin \: \alpha \: sin \: 3 \alpha - {sin}^{2} \: 3 \alpha ) \\ \\ \rm = \frac{1}{4}. \{ {cos}^{2} 3 \alpha - {sin}^{2}3 \alpha + 3 (cos \: \alpha \: cos \: 3 \alpha + sin \: \alpha \: sin \: 3 \alpha ) \} \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc} \orange{{\boxed{\begin{array}{cc}\text{we \: know \: that} \\ \\ \to \: \rm \: {cos}^{2} \: x - {sin}^{2} \: x = cos \: 2x \\ \\ \to \: \rm \: cos \: x \: cos \: y + sin \: x \: sin \: y = cos(x - y) \\ \end{array}}}} \\ \pink{ \text{ apply \: this}} \\ \\ \rm = \frac{1}{4}. \{cos \: 2.3 \alpha + 3cos(3x - x) \} \\ \\ \rm = \frac{1}{4}(cos \: 3.2 \alpha + 3cos \: 2 \alpha ) \\ \\ \rm= \frac{1}{4} \times 4 \: {cos}^{3} \: 2 \alpha \\ \\ \rm = {cos}^{3} \: 3 \alpha \\ \\ \rm = R.H.S\end{array}}}}$

Hence, proved

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Assalamu A'laikum,

Good morning... $\ddot\smile$

Tomader ki 1st year final exam a short syllabus er upor exam hobe naki syllabus komano hoini ?