Question

what is the value of sin60°- sin 30°?​

Answer 1

$\implies \sf\sin60 \degree - \sin30 \degree$

We know that :-

$\boxed{ \bf \sin60 \degree = \dfrac{ \sqrt{3} }{2} }$

$\boxed{ \bf \sin30 \degree = \dfrac{ 1 }{2} }$

$\implies \sf\sin60 \degree - \sin30 \degree = \dfrac{ \sqrt{3} }{2} - \dfrac{ 1 }{2}$

LCM= 2

$\implies \sf\sin60 \degree - \sin30 \degree = \dfrac{ \sqrt{3} - 1}{2}$

Answer :

$\bf \dfrac{ \sqrt{3} - 1}{2}$

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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

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

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\huge\boxed{\sf{\sin \left(60^{\circ \:}\right)-\sin \left(30^{\circ \:}\right)}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\sin \left(60^{\circ \:}\right)-\sin \left(30^{\circ \:}\right)=\dfrac{\sqrt{3}-1}{2}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(60^{\circ \:}\right)=\dfrac{\sqrt{3}}{2}$

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

$=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}$

$\mathrm{Apply\:rule}\:\dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$\large\boxed{\sf{=\dfrac{\sqrt{3}-1}{2}}}$