Question

Q. 2] Find the values of
1) Sin 2 60°- Sin 90°



Please help me.​

Answer 1

Answer:

sin 2 60- sin 90

putting values of sin 60and 90

2*√3/2 - 1

√3- 1

thats the answer

Answer 2

Given : Expression = sin² 60⁰ - sin 90⁰

Need To Evaluate : sin² 60⁰ - sin 90⁰

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❍ Trigonometric Ratios for standard angles :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

⠀⠀⠀⠀⠀By Seeing Table of Trigonometry Ratios for angles :

• $\sin 60\degree = \dfrac {\sqrt{3}}{2}$
• $\sin 90\degree = 1$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Expression = sin² 60⁰ - sin 90⁰

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad :\implies \sf{ \sin^2 60\degree - \sin 90\degree}$

$\qquad :\implies \sf{ \bigg( \dfrac{\sqrt{3}}{2} \bigg)^2 - 1 }$

$\qquad :\implies \sf{ \dfrac{3}{4} - 1 }$

$\qquad :\implies \sf{ \dfrac{3-4}{4} }$

$\qquad :\implies \bf{Answer\:= \dfrac{-1}{4} }$

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