[tex]{\huge{\boxed{\sf{\green{Question}}}}}[/tex]

1.)Show that 2 [tex]Sin^{2}[/tex]β + 4 cos (α + β) sin α sin β + cos

2 thoughts on “[tex]{\huge{\boxed{\sf{\green{Question}}}}}[/tex]<br /><br /> 1.)Show that 2 [tex]Sin^{2}[/tex]β + 4 cos (α + β) sin α sin β + cos”

1. 1. ANSWER:

   To Prove:

   • 2sin²β + 4cos(α + β) × sin α × sin β + cos 2(α + β) = cos 2α.

   Proof:

   [tex]\text{We need to prove that,}\\\\:\longrightarrow2\sin^2\beta + 4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos2(\alpha+\beta)=\cos2\alpha\\\\\text{Taking LHS,}\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos2(\alpha+\beta)\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos(2\alpha+2\beta)\\\\\text{We know that,}\\\\:\hookrightarrow\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi[/tex]

   [tex]\text{So,}\\\\:\implies2\sin^2\beta+4\cos(\alpha+\beta)\times\sin\alpha\sin\beta+\cos(2\alpha+2\beta)\\\\:\implies2\sin^2\beta+4(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\times\sin\alpha\sin\beta+(\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta)[/tex]

   [tex]:\implies2\sin^2\beta+4\cos\alpha\cos\beta\sin\alpha\sin\beta-4\sin^2\alpha\sin^2\beta+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta\\\\:\implies2\sin^2\beta+(2\sin\alpha\cos\alpha)(2\sin\beta\cos\beta)-4\sin^2\alpha\sin^2\beta+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta[/tex]

   [tex]\text{We know that,}\\\\:\hookrightarrow2\sin\theta\cos\theta=\sin2\theta\\\\\text{So,}\\\\:\implies2\sin^2\beta+(2\sin\alpha\cos\alpha)(2\sin\beta\cos\beta)-4\sin^2\alpha\sin^2\beta+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta\\\\:\implies2\sin^2\beta+\sin2\alpha\sin2\beta-(2\sin^2\alpha)(2\sin^2\beta)+\cos2\alpha\cos2\beta-\sin2\alpha\sin2\beta[/tex]

   [tex]\text{On canceling $(+\sin2\alpha\sin2\beta)$ and $(-\sin2\alpha\sin2\beta)$,}\\\\:\implies2\sin^2\beta-(2\sin^2\alpha)(2\sin^2\beta)+\cos2\alpha\cos2\beta\\\\\text{We know that,}\\\\:\hookrightarrow2\sin^2\theta=1-\cos2\theta\\\\\text{So,}\\\\:\implies(2\sin^2\beta)-(2\sin^2\alpha)(2\sin^2\beta)+\cos2\alpha\cos2\beta\\\\:\implies(1-\cos2\beta)-(1-\cos2\alpha)(1-\cos2\beta)+\cos2\alpha\cos2\beta\\\\:\implies1-\cos2\beta-(1-\cos2\beta-\cos2\alpha+\cos2\alpha\cos2\beta)+\cos2\alpha\cos2\beta[/tex]

   [tex]:\implies1-\cos2\beta-1+\cos2\beta+\cos2\alpha-\cos2\alpha\cos2\beta+\cos2\alpha\cos2\beta\\\\\text{On rearranging,}\\\\:\implies(1-1)+(\cos2\beta-\cos2\beta)+\cos2\alpha+(\cos2\alpha\cos2\beta-\cos2\alpha\cos2\beta)\\\\:\implies0+0+\cos2\alpha+0\\\\\bf{:\implies \cos2\alpha=RHS}\\\\\text{\bf{HENCE PROVED!!!}}[/tex]

   Formulae Used:

   [tex]:\hookrightarrow1)\:\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi\\\\:\hookrightarrow2)\:2\sin\theta\cos\theta=\sin2\theta\\\\:\hookrightarrow3)\:2\sin^2\theta=1-\cos2\theta[/tex]

   [tex]\\\\\\[/tex]

   2. ANSWER:

   Given:

   • Revolutions per minute by a wheel = 360

   To Find:

   • Radians covered in 1 second.

   Solution:

   [tex]\text{We are given that,}\\\\:\longrightarrow\text{Number of revolutions in 1 minute by a wheel}=360\\\\\text{As, 1minute = 60 seconds,}\\\\:\implies\text{Number of revolutions in 60 second}=360\\\\\text{So,}\\\\:\implies\text{Number of revolutions in 1 second}=\dfrac{360}{60}\\\\:\implies\text{Number of revolutions in 1 second}=6\\\\\text{We know that,}\\\\:\hookrightarrow\text{Angle made in 1 revolution}=2\pi^c\\\\\text{So,}[/tex]

   [tex]:\implies\text{Angle made in 6 revolutions}=6 \times2\pi^c\\\\:\implies\text{Angle made in 6 revolutions}=12\pi^c\\\\\text{Hence,}\\\\\bf{:\implies\text{\bf{Radians covered in 1 second(6 revolutions)}}=12\pi}[/tex]

   Formula Used:

   • 1 minute = 60 seconds
   • Angle made in 1 revolution = 2π radian

   [tex]\\\\\\[/tex]

   3. ANSWER:

   Given:

   • Angle = 60°
   • Arc length = 37.4cm

   To Find:

   • Radius of the circle(r)

   Solution:

   [tex]\text{We are given that,}\\\\:\longrightarrow \text{Length of arc}(l)=37.4cm\\\\\text{And,}\\\\:\longrightarrow\text{Angle}(\theta)=60^{\circ}\\\\\text{We know that,}\\\\:\hookrightarrow\text{Radian Measure}=\dfrac{\pi}{180}\times\text{Degree measure}\\\\\text{So,}\\\\:\implies\text{Angle}(\theta)=\left(\dfrac{\pi}{180}\times60\right)^c\\\\:\implies\text{Angle}(\theta)=\dfrac{\pi}{3}^c=\dfrac{22}{7\times3}^c=\dfrac{22}{21}^c\\\\\text{We know that,}[/tex]

   [tex]:\hookrightarrow\text{Length of arc = radius $\times$ angle}\\\\:\hookrightarrow l = r\theta\\\\\text{So,}\\\\:\implies l = r\theta\\\\:\implies r=\dfrac{l}{\theta}\\\\:\implies r=\dfrac{37.4}{\frac{22}{21}}\\\\:\implies r=\dfrac{37.4\!\!\!\!\!\!/^{\:\:\:1.7}\times21}{22\!\!\!\!\!/_{\:\:1}}\\\\:\implies r=1.7\times21\\\\\bf{:\implies r=35.7cm}\\\\\text{\bf{Hence, the radius of the circle is 35.7cm}}[/tex]

   Formulae Used:

   [tex]:\hookrightarrow1)\:\text{Radian Measure}=\dfrac{\pi}{180}\times\text{Degree measure}\\\\:\hookrightarrow2)\:\text{Length of arc = radius $\times$ angle}[/tex]

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2. [tex]\Huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}}{\color{pink}{:}}}}}[/tex]

   [2]A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? In one complete revolution, the wheel turns an angle of 2π radian. Thus, in one second, the wheel turns an angle of 12π radian.

   Hᴏᴘᴇ ᴛʜɪs ʜᴇʟᴘs ᴜᴜʜʜ!!❥︎Kʜᴜsʜɪ

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