Question

value of 3 sin 30 - 4 sin cube 30+4cos cube 30-3 cos 30​

Answer 1

We have to find the value of :-

$3 \sin(30\degree) - 4 { \sin}^{3} (30\degree) + 4 { \cos}^{3} (30\degree) - 3 \cos(30\degree)$

we know that :

$\implies\sin(30\degree) = \dfrac{1}{2}$

$\implies\cos(30\degree) = \dfrac{ \sqrt{3} }{2}$

Now put above values in the given expression :-

$\implies3 \sin(30\degree) - 4 { \sin}^{3} (30\degree) + 4 { \cos}^{3} (30\degree) - 3 \cos(30\degree)$

$\implies(3 \times \dfrac{1}{2} ) - 4 {( \frac{1}{2} )}^{3} + 4 { (\frac{ \sqrt{3} }{2}) }^{3} - (3 \times \frac{ \sqrt{3} }{2} )$

$\implies \dfrac{3}{2} - (4 \times { \dfrac{1}{8} } )+ 4 { (\dfrac{ 3\sqrt{3} }{8}) } - (\dfrac{ 3\sqrt{3} }{2} )$

$\implies \dfrac{3}{2} - ( { \dfrac{1}{2} } )+ { (\dfrac{ 3\sqrt{3} }{2}) } - ( \dfrac{ 3\sqrt{3} }{2} )$

$\implies \dfrac{3}{2} - { \dfrac{1}{2} } + 0$

LCM = 2

$\implies \dfrac{3 - 1}{2} + 0$

$\implies \dfrac{2}{2}$

Cancel out 2 from numerator and denominator :-

$\implies 1$

Answer : 1

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$\large\bf\red{Additional-Information}$

1. Cosθ = base / hypotenuse

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ

$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\boxed{\bf{3\sin \left(30^{\circ \:}\right)-4\sin ^3\left(30^{\circ \:}\right)+4\cos ^3\left(30^{\circ \:}\right)-3\cos \left(30^{\circ \:}\right)}}$

♣ ᴀɴꜱᴡᴇʀ :

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(30^{\circ \:}\right)=\frac{1}{2}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(30^{\circ \:}\right)=\frac{\sqrt{3}}{2}$

$\sf{=3\cdot \dfrac{1}{2}-4\left(\dfrac{1}{2}\right)^3+4\left(\dfrac{\sqrt{3}}{2}\right)^3-3\cdot \dfrac{\sqrt{3}}{2}}$

$\sf{3\cdot \dfrac{1}{2}-4\left(\dfrac{1}{2}\right)^3+4\left(\dfrac{\sqrt{3}}{2}\right)^3-3\cdot \dfrac{\sqrt{3}}{2}=1}$

$\large\boxed{\bf{3\sin \left(30^{\circ \:}\right)-4\sin ^3\left(30^{\circ \:}\right)+4\cos ^3\left(30^{\circ \:}\right)-3\cos \left(30^{\circ \:}\right)=1}}$